# Units 1-3 Review

Motion in Two-Dimension

1. If an object has an acceleration of 0 m/s2, then one can be sure that the object is not

 a. moving b. changing position c. changing velocity

An object with zero acceleration could be at rest and staying at rest (in which case velocity is a constant value of 0 m/s) or in motion and staying in motion (in which velocity is a constant value). In either case, the velocity is NOT changing.

Acceleration (14 seconds)

2. Which one of the following is NOT consistent with a car which is accelerating?

1. moving with an increasing speed.
2. moving with a decreasing speed.
3. moving with a high speed.
4. changing direction.

An accelerating object is an object which is changing its velocity. A velocity change can result from a change in either speed (as in choice a and b) or direction (as in case d). An object moving at a high speed is not necessarily accelerating; if its speed is constant, then it is not accelerating.

Acceleration (14 seconds)

3. Barry Sanders is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. This is evidence that

1. he is accelerating
2. he is covering a greater distance in each consecutive second.
3. he is moving with a constant speed (on average).

Barry Sanders is covering a constant distance of 10 yards every second. For this reason his speed is constant and his acceleration is 0 m/s/s (assuming he is not changing his direction).

Acceleration (14 seconds) | Speed vs. Velocity (8 seconds)

4. Barry Sanders is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. What is Barry's acceleration?

Barry Sanders is covering a constant distance of 10 yards every second. For this reason his speed is constant and his acceleration is 0 m/s/s (assuming he is not changing his direction). Note that if there was an acceleration, it would be difficult to determine its value from mere position-time data.

Acceleration (14 seconds) | Speed vs. Velocity (8 seconds)

5. If an object is moving eastward and slowing down, then the direction of its acceleration vector is

 a. eastward b. westward c. neither d. not enough info to tell

When an object slows down, the the direction of the acceleration vector is in the direction opposite to which the object moves. An eastward moving object which is slowing down would experience a westward-directed net force which causes a westward acceleration.

Acceleration (14 seconds)

6. Which one of the following quantities is NOT a vector?

 a. 10 mi/hr, east b. 10 mi/hr/sec, west c. 35 m/s, north d. 20 m/s

A vector quantity has both magnitude and direction. Only choice D is listed with magnitude alone.

Scalars vs. Vectors (3 seconds)

7. Which one of the following statements is NOT true of a free-falling object? An object in a state of free fall

 a. falls with a constant speed of -10 m/s. b. falls with a acceleration of -10 m/s/s. c. falls under the sole influence of gravity. d. falls with an acceleration of constant magnitude.

As an object free-falls, its velocity (and also its speed) changes by approximately 10 m/s every second. There is no way that it can be concluded that the speed is constant. Do not confuse the idea of the rate at which the speed changes (also known as acceleration) - which is a constant change of 10 m/s during each second - with the actual speed value - which is continuously changing in any accelerated type motion.

Introduction to Free Fall (4 seconds) | The Acceleration of Gravity (6 seconds)

8. What is the acceleration of a car that maintains a constant velocity of 100 km/hr for 10 seconds?

 a. 0 b. 10 km/hr/s c. 10 km/s/s d. 1000 km/hr/s

If the velocity is constant, then the acceleration is 0. If you selected b or c, perhaps you are erroneously using the equation a=v/t. There is no such equation. There is an equation a = (velocity change)/time. In this case, the velocity change = 0 mkm/hr.

Acceleration (14 seconds)

9. As an object freely falls, its

 a. speed increases b. acceleration increases c. both of these d. none of these

As an object free-falls, its velocity (and also its speed) changes by approximately 10 m/s every second. This means that the acceleration is a constant value of 10 m/s. An object has a changing speed (or velocity) and a constant acceleration if the speed changes by the same amount (a "constant amount") in each consecutive second of its motion.

Introduction to Free Fall (4 seconds) | The Acceleration of Gravity (6 seconds)

10. A ball is thrown into the air at some angle between 0 and 90 degrees. At the very top of the ball's path, its velocity is _______.

 a. entirely vertical b. entirely horizontal c. both vertical and horizontal d. not enough information given to know.

As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. At its peak, its vertical velocity becomes 0 m/s. At this instant in time, the velocity is entirely horizontal; there is no vertical component to the velocity.

Projectiles: Horizontal and Vertical Components of Velocity (11 seconds) | Projectile Animation (15 seconds)

11. A ball is thrown into the air at some angle between 0 and 90 degrees. At the very top of the ball's path, its acceleration is _______.

 a. entirely vertical b. entirely horizontal c. both vertical and horizontal d. not enough information given to know.

As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.

12. A ball is thrown into the air at some angle between 0 and 90 degrees. Neglecting air resistance, at the very top of the ball's path, the net force acting upon it is _______.

 a. entirely vertical b. entirely horizontal c. both vertical and horizontal d. not enough information given to know.

A projectile is an object upon which the only force is gravity. Since no other forces act upon the object, the net force would be downward.

13. At what point in its path is the horizontal component of the velocity of a projectile the smallest?

 a. when it is thrown. b. half-way to the top. c. at the top. d. as it nears the top. e. it is the same throughout the path.

As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. For this reason, E is the answer.

Projectiles: Horizontal and Vertical Components of Velocity (11 seconds) | Projectile Animation (12 seconds)

14. At what point in its path is the vertical component of the velocity of a projectile the smallest?

 a. when it is thrown. b. half-way to the top. c. at the top. d. as it nears the ground. e. it is the same throughout the path.

As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. During the upward portion of its trajectory, the vy continuosly decreases until it becomes 0 m/s at the peak.

Projectiles: Horizontal and Vertical Components of Velocity (11 seconds) | Projectile Animation (15 seconds)

15. An airplane that flies at 100 km/h in a 100 km/h hurricane crosswind has a ground speed of

 a. 0 km/h b. 100 km/h c. 141 km/h d. 200 km/h

When an object such as a plane or a boat moves within a medium which is moving relative to the ground, the speed of the plane or boat (as measured by its speedometer) will not be the same as the speed as measured by a person on the ground (ground speed). The ground speed (i.e., resultant velocity produced by the combination of the plane's speed and the wind speed) can be determined by adding the plane speed and wind speed as vectors. In this case, the two vectors are at right angles, so the resultant velocity can be determined using Pythagorean theorem.

R2 = (100 km/hr)2 + (100 km/hr)2

R = SQRT( (100 km/hr)2 + (100 km/hr)2 )

R = 141 km/hr

16. An airplane travels at 141 km/h toward the northeast (45 degrees). What is its component velocity due north?

 a. 41 km/h b. 100 km/h c. 110 km/h d. 141 km/h

To determine the component of a vector in a given direction, vector resolution principles must be applied. This velocity vector is directed at 45 degrees and has magnitude of 141 km/hr. The vertical component (northward) of this vector can be found using the sine function.

vy = v * sine(theta)

vy = (141 km/hr)*sine(45 degrees) = 100 km/hr

Vector Components (15 seconds)

17. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity

 a. decreases. b. remains constant c. increases

Once the ball leaves the table's edge, it becomes a projectile. As it falls, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.

Projectiles: Horizontal and Vertical Components of Velocity (11 seconds) | Projectile Animation (12 seconds)

18. A bullet fired horizontally hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it would have hit the ground in (assume no air resistance)

 a. less than 0.5 s. b. more than 0.5 s. c. 0.5 s.

Once the bullet leaves the table's edge, it becomes a projectile (assuming no air resistance). As it falls, its horizontal velocity remains constant while its vertical velocity decreases. The force of gravity acts upon the bullet to cause its downward acceleration. The motion of the bullet in the downward direction is independent of the motion in the horizontal direction. That is to say, any alteration in a horizontal aspect of its motion will not effect the motion in the vertical direction. The time to fall vertically to the ground is not effected by the horizontal speed of the projectile.

19. A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. What other projection angle for the same speed would produce the same down-range distance?

 a. 30 degrees. b. 45 degrees. c. 50 degrees. d. 75 degrees e. 90 degrees.

For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Any two launch angles which are separated from 45 degrees by the same amount (for example, 40 degrees and 50 degrees, 30 degrees and 60 degrees and 15 degrees and 75 degrees) will provide the same horizontal displacement.

20. Two projectiles are fired at equal speeds but different angles. One is fired at angle of 30 degrees and the other at 60 degrees. The projectile to hit the ground first will be the one fired at (neglect air resistance)

 a. 30 degrees b. 60 degrees c. both hit at the same time

For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Launch angles greater than 45 degrees result in large vertical components of velocity; these stay in the air longer and rise to higher heights. Launch angles less than 45 degrees result in small vertical comonents of velocity; these do not rise as high and end up falling to the ground in small times.

Maximum Range Animation (13 seconds)

21. Express the direction of each of the following vectors in the diagram below.

 A: ______ B: ______ C: ______ D: ______ E: ______ F: ______

All measurements are approximate and given in degrees

 A: 0 B: 90 C: 135 D: 205 E: 270 F: 340

We have been using the following convention for expressing the direction of a vector: a vector's direction is expressed as the counter-clockwise angle of rotation of that vector from due East.

Vectors and Direction (11 seconds)

22. In the following diagrams, two vectors are being added and the resultant is drawn. For each diagram, identify which vector is the resultant and write the equation (e.g., A + B = C).

 a. The resultant is vector _____. The equation is ____ + ____ = _____ b. The resultant is vector _____. The equation is ____ + ____ = _____ c. The resultant is vector _____. The equation is ____ + ____ = _____ d. The resultant is vector _____. The equation is ____ + ____ = _____

Vectors are added head-to-tail (just like the way elephants walk around in a circus). The resultant of two or more vectors is drawn from the tail of the first vector to the head of the last vector. Thus, the resultant will be the vector which has its tail next to another tail and its head next to another head. This principle will allow you to determine the following answers:

 a. The resultant is vector B. The equation is C + A = B b. The resultant is vector B. The equation is A + C = B c. The resultant is vector A. The equation is C + B = A d. The resultant is vector C. The equation is A + B = C

Vector Addition (20 seconds) | Resultants (6 seconds)

23. A hiker's motion can be described by the following three displacement vectors.

22 km, 45 degrees + 16 km, 135 degrees + 12 km, 270 degrees

1. What is the distance walked by the hiker?
2. What is the resulting displacement of the hiker?

Vectors must be added head-to-tail (just like the way elephants walk around in a circus). The resultant of two or more vectors is drawn from the tail of the first vector to the head of the last vector. A scale must first be designated and then each vector is drawn in consecutive fashion, with the tail of the next vector starting at the head of the most recent vector. See diagram below.

The magnitude and direction of R can be measured using a ruler and protractor.

R = 15.5 km. 74 degrees

Since displacement is a vector, the vector sum of the three displacement vectors provides the resultant displacement of the hiker. The algebraic sum of the distances for each leg of the trip provides the total distance hiked.

 a. distance walked = 50 km b. resultant displacement = 15.5 km. 74 degrees

Vector Addition (20 seconds) | Resultants (6 seconds) | Distance vs. Displacement (7 seconds)

Consider the following statements in answering the next three questions.

1. The position of the object remains constant.
2. The velocity of the object remains constant.
3. The position of the object is changing.
4. The velocity of the object is changing.
5. The velocity of the object is 0 m/s.
6. The acceleration of the object is 0 m/s2.
7. The resultant of all the individual forces acting upon the object is 0 N.
8. The net force acting upon the object is 0 N.
9. All individual forces acting upon the object are equal.
10. The individual forces acting upon the object balance each other.

24. If an object is known to be at equilibrium, then which of the following statements MUST be true?

25. If an object is known to be at equilibrium, then which of the following statements MUST NOT be true?

26. If an object is known to be at equilibrium, then which of the following statements COULD be true?

If an object is at equilibrium, then all the forces acting upon the object would balance each other. That is, the net force acting upon the object would be 0 N. If the forces were added together as vectors in a head-to-tail fashion, then the resultant would be 0 N. If the net force is 0 N, then the acceleration of the object would be 0 m/s/s. That means the velocity of the object (whether it be 0 m/s or not 0 m/s) must be a constant value. So an object which is at rest (v = 0 m/s) would stay at rest in the same position; but an object which is moving would remain moving with that very velocity. So an object at equilibrium could be moving with a constant velocity (and change its position) or could be at rest (and remain at a constant position). For these reasons, the following answers exist for questions #24-26.

 24. B, F, G, H, and J 25. D 26. A, C, E, and I

Equilibrium and Statics (23 seconds) | Acceleration (14 seconds)

27. What is the maximum resultant force resulting from a 3-N force and an 8-N force? ______

28. What is the minimum resultant force resulting from a 3-N force and an 8-N force? ______

When two forces are added, the maximum resultant is the result of adding them in the same direction and the minimum resultant is the result of adding them in the opposite direction. Subsequently, the answers to #27 and #28 are:

 27. maximum = 11 N 28. minimum = 5 N

29. A boat heads straight across a river which is 100 meters wide. For the following two combinations of boat velocities and current velocities, determine the resultant velocity, the time required to cross the river, and the distance traveled downstream.

 a. Given: Boat velocity = 10 m/s, East River velocity = 4 m/s, North Calculate: Resultant Vel. (mag. & dir'n): ________ Time to cross river: _________ Distance traveled downstream: _______ b. Given: Boat velocity = 8 m/s, East River velocity = 5 m/s, South Calculate: Resultant Vel. (mag. & dir'n): ________ Time to cross river: _________ Distance traveled downstream: _______

For each of these problems, construct a triangle (as done in class) to add the two given vectors in a head-to-tail method. Refer to pp. 11-14 of your Packet (River Boat Simulation) if you need further help.

 a. Resultant Vel. (mag. & dir'n): R2 = (10 m/s)2 + (4 m/s)2 R = SQRT ( (10 m/s)2 + (4 m/s)2 ) R = SQRT(116 m2/s2) = 10.8 m/s dir'n = invtan(4/10) = 22 degrees Time to cross river: d = v*t ---> t = d/v t = (100 m)/(10 m/s) = 10 s Distance traveled downstream: d = v*t = (4 m/s)*(10 s) = 40 m b. Resultant Vel. (mag. & dir'n): R2 = (8 m/s)2 + (5 m/s)2 R = SQRT ( (8 m/s)2 + (5 m/s)2 ) R = SQRT(89 m2/s2) = 9.43 m/s dir'n = 360 deg. - invtan(5/8) = 328 degrees Time to cross river: d = v*t ---> t = d/v t = (100 m)/(8 m/s) = 12.5 s Distance traveled downstream: d = v*t = (5 m/s)*(12.5 s) = 62.5 m

30. The diagram at the right depicts a horizontally-launched projectile leaving a cliff of height y with a horizontal velocity (vix) and landing a distance x from the base of the cliff. Express your understanding of projectile kinematics by filling in the blanks in the table below.

 vix (m/s) y (m) t (s) x (m) a. 15 m/s 20 m ______ ______ b. 15 m/s ______ 3.0 s ______ c. ______ 45 m ______ 45 m d. ______ ______ 2.50 s 30 m e. ______ 74.0 m ______ 66 m

The solutions to all five of these projectile problems involve the use of kinematic equations and an appropriate problem-solving strategy. The kinematic equations and their use in projectile problems are listed and discussed elsewhere. In each of these problems, it is known that ax = 0 m/s/s, ay = -10 m/s/s, and viy = 0 m/s. When these three knowns are combined with the other given knowns the following answers are obtained:

 Answers Method a. t = 2 s and x = 30 m Use y, viy, and ay to calculate t; then use t, vix and ax to calculate x. b. y = 45 m and x = 45 m Use t, vix and ax to calculate x; and use t, viy, and ay to calculate y. c. vix = 15.0 m/s and t = 3.0 s Use y, viy, and ay to calculate t; then use t, x and ax to calculate vix. d. vix = 12.0 m/s and y = 31.3 m Use t, x and ax to calculate vix; and use t, viy, and ay to calculate y. e. vix = 17.2 m/s and t = 3.85 s Use y, viy, and ay to calculate t; then use t, x and ax to calculate vix.

31. The launch velocity and angle is given for three different projectiles. Use trigonometric functions to resolve the velocity vectors into horizontal and vertical velocity components. Then use kinematic equations to determine the time that the projectile is in the air, the height to which it travels (when it is at its peak), and the horizontal distance that it travels.

 a. Given: Launch Vel. = 30 m/s Launch angle = 30 degrees b. Given: Launch Vel. = 30 m/s Launch angle = 45 degrees c. Given: Launch Vel. = 30 m/s Launch angle = 50 degrees Calculate: vix = __________ viy = ___________ tup = ___________ ttotal = ___________ y at peak = ___________ x = ___________ Calculate: vix = __________ viy = ___________ tup = ___________ ttotal = ___________ y at peak = ___________ x = ___________ Calculate: vix = __________ viy = ___________ tup = ___________ ttotal = ___________ y at peak = ___________ x = ___________

The solutions to all three of these non-horizontally launched projectile problems involve the use of kinematic equations and an appropriate problem-solving strategy. The kinematic equations and their use in projectile problems are listed and discussed elsewhere. In each of these problems, it is known that ax = 0 m/s/s and ay = -10 m/s/s. The values of vix and viy can be determined using trigonometric functions:

 vix = vi * cos(theta) viy = vi * sin(theta)

Once vix and viy are known, the other unknowns can be calculated. The time up to the peak (tup) can be determined using the equation

vfy = viy + ay *t

where the vfy = 0 m/s (there is no vertical velocity for a projectile when its at its peak) and ay = -10 m/s/s. Once tup is known, the ttotal (time to travel the entire trajectory -both up and down) can be determined by doubling the tup. The horizontal displacement of the projectile (x) can be computed in the usual way using the equation

x = vix*t + 0.5*ax*t2

where t is the ttotal value, ax = 0 m/s/s and vix was the first value calculated (using the trigonometric functions). Finally, the y at the peak (i.e., the peak height) can be calculated using the equation

y = viy*t + 0.5*ay*t2

where t is the tup value, ay = -10 m/s/s and viy was one of the first values calculated (using the trigonometric function). The t value in the equation is tup because the peak height is reached when the projectile has traveled for one-half of its total time; tup is that time. This method will yield the following answers:

 a. Answers: vix = 26.0 m/s viy = 15 m/s tup = 1.5 s ttotal = 3.0 s y at peak = 11.3 m x = 77.9 m b. Answers: vix = 21.2 m/s viy = 21.2 m/s tup = 2.12 ttotal = 4.24 y at peak = 22.5 m x = 89.9 m c. Answers: vix = 19.3 m/s viy = 23.0 m/s tup = 2.3 s ttotal = 4.6 s y at peak = 26.5 m x = 88.8 m

32. What angle above gives the projectile the maximum range (i.e., maximum horizontal displacement)?

Observe that the maximum range (x) in the above three calculations results when the angle of launch is 45-degrees. This is always the case for situations where air resistance is negligible.

Maximum Range Animation (13 seconds)

33. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6-meter high building. Determine the horizontal displacement of the projectile.

A problem such as this is almost a sure bet on the test. This horizontally-launched projectile problem can be (and should be) solved in the same manner as the solution to #30 above. While #30 is broken down for you into nice steps, this problem is not so user-friendly. It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations:

 x = ??? vix = 12.0 m/s ax = 0 m/s/s (true for all projectiles) y = -24.6 m (- means falling down) viy = 0.0 m/s (its launched horizontally) ay = -10 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time.

y = viy*t + 0.5*ay*t2

Plugging a chugging the above values into this equation yields a time of 2.22 seconds. Now the t value can be combined with the vix and ax value and used in an x-equation

x = vix*t + 0.5*ax*t2

to yield the answer 26.6 m. More examples and discussion of these types of projectile problems are discussed elsewhere.

34. A projectile is launched with an initial speed of 21.8 m/s at an angle of 35-degrees above the horizontal. Determine the horizontal displacement of the projectile.

A problem such as this is also almost a sure bet on the test. This non-horizontally-launched projectile problem can be (and should be) solved in the same manner as the solution to #31 above. While #31 is broken down for you into nicely-structured steps, this problem is not so user-friendly. It is strongly recommended that you begin by resolving the initial velocity and angle into initial velocity components using the equations:

 vix = vi * cos(theta) viy = vi * sin(theta)

This yields values of vix = 17.9 m/s and viy = 12.5 m/s. Once done, list the known values for each of the variables in the kinematic equations:

 x = ??? vix = 17.9 m/s (from trig. function) ax = 0 m/s/s (true for all projectiles) y = 0 m (it rises and falls to original height) viy = 12.5 m/s (from trig. function) ay = -10 m/s/s (true for all projectiles)

Since three pieces of y-information are now known, a y-equation can be employed to find the time. Either use

y = viy*t + 0.5*ay*t2

in which case there will be two solutions: t = 0 s and t = 2.50 s. These two solutions to the equation indicate that the time is 0 s when the vertical displacement (y) is 0 m. This is true before being launched (t = 0 s) and the instant it lands (t=2.50 s). The latter of the two solutions can be used to determine the horizontal displacement (x). Use the equation:

x = vix*t + 0.5*ax*t2

where t is 2.50 s, ax = 0 m/s/s and vix was the first value calculated (using the trigonometric functions). Plugging a chugging the above values into this equation yields the answer of 44.8 m. More examples and discussion of these types of projectile problems are discussed elsewhere.

35. The signs in the two diagrams below hang at equilibrium by means of two cables which are arranged symmetrically. The tension in both cables and the angle between the cables is indicated. Use trigonometric functions and force principles to determine the mass and the weight of the following three signs.

 a. b. mass = __________ Weight = _________ mass = __________ Weight = _________

This problem and the next represent typical equilibrium problems - one of which would likely be a sure bet on the test (except with different numbers of course). The key to these problems is to conduct a free-body analysis using trigonometric functions in an effort to balance all forces acting upon the sign. Since the ropes are at angles, the analysis becomes simplified by resolving the tension force into x- and y-components. First find the angle which the rope makes with the horizontal (45 degrees in a. and 60 degrees in b.) using the principle that all three angles must add to 180-degrees. The Fx and Fy values can now be calculated using SOH CAH TOA. The important value of the two values is Fy since Fy in each of the two cables must provide sufficient upward force to balance the downward force of gravity on the sign. The force of gravity (a.k.a. "weight") then is the sum of the two Fy values. The mass is merely the force of gravity divided by the acceleration of gravity (g) which is approximated as 10 m/s/s. The answers are shown below.

 a. angle with horizontal = 45 degrees Fy = (100 N)*sin 45 = 70.7 N Fgrav = Weight = 2 * Fy = 141 N m = (Fgrav)/g = 14.1 kg b. angle with horizontal = 60 degrees Fy = (80 N)*sin 60 = 69.3 N Fgrav = Weight = 2 * Fy = 139 N m = (Fgrav)/g = 13.9 kg

Several additional examples and discussion of equilibrium problems are provided elsewhere at this site.

Resolution of Forces (22 seconds) | Equilibrium and Statics (23 seconds)

36. The signs in the two diagrams below hang at equilibrium by means of two cables which are arranged symmetrically. The mass of the sign and the angle between the cables is indicated. Use trigonometric functions and force principles to determine the tension in the cables.

 a. m = 5 kg b. m = 8 kg Tension = __________ Tension = __________

This problem (like the prior problem) represents a typical equilibrium problem - one of which would likely be a sure bet on the test (except with different numbers of course). The key to these problems is to conduct a free-body analysis using trigonometric functions in an effort to balance all forces acting upon the sign. Begin by determining the weight of the signs using the equation

Weight = Fgrav = m*g where g = 10 m/s/s (approx.)

Since the downward force of gravity must be balanced by the vertical component of the Ftens values, the each individual cable must pull upward with a force equal to one-half the Fgrav value. When the upward pull of the two cables are combined, it will balance the downward pull of gravity. So dividing the Fgrav value by two yields the vertical component (Fy) of force in the cables. The Ftens value can now be calculated using SOH CAH TOA. Since the angle and the length of the side opposite (Fy) are known, the sine function is used. The answers are shown below.

 a. angle with horizontal = 30 degrees Fgrav = m*g = 50 N Fy = Fgrav /2 = 25 N sin(30) = Fy/Ftens = (25 N)/Ftens Do some good algebra to rearrange the above equation into the form of Ftens = (25 N)/sin(30) Ftens = 50 N b. angle with horizontal = 25 degrees Fgrav = m*g = 80 N Fy = Fgrav /2 = 40 N sin(25) = Fy/Ftens = (40 N)/Ftens Do some good algebra to rearrange the above equation into the form of Ftens = (40 N)/sin(25) Ftens = 95 N

Several additional examples and discussion of equilibrium problems are provided elsewhere at this site.

Resolution of Forces (22 seconds) | Equilibrium and Statics (23 seconds)

37. The following diagrams depict a force being applied at an angle in order to accelerate an object across a rough surface. For each case, determine the values of all missing quantities (Fgrav, Fnorm, Ffrict, Fnet, and acceleration).

 a. m = 12 kg Fnet = __________, ________ (dir'n) a = __________, ________ (dir'n) b. m = 8 kg Fnet = __________, ________ (dir'n) a = __________, ________ (dir'n)

This problem would be no different than problems done in Unit 2 if the applied force were merely horizontal. Unfortunately the problem has a slightly higher degree of complexity. The complexity can be partly removed by using SOH CAH TOA to resolve the applied force into horizontal and vertical components. Once resolved, the Fapp can be disregarded since its components have been substituted into its place. A free-body analysis must now be conducted to determine the net force and the acceleration. Begin by computing Fgrav using the equation

Fgrav = m*g where g = 10 m/s/s (approx.)

This analysis procedes by determining Fnorm using the principle that the vertical forces must balance if there is no vertical acceleration (the acceleration is horizontally across the rough surface). So

Fnorm + Fy = Fgrav

Once Fnorm is determined, the Ffrict can be calculated using the coefficient of friction (mu) value and the following equation:

Ffrict = mu*Fnorm

Now the net force can be determined by adding all the forces as vectors. The vertical forces cancel; yet Fx (to the right) is greater than Ffrict (to the left). Subtract the smaller force from the larger force to determine Fnet. Then use the equation

Fnet = m*a

to determine the acceleration. The following answers are found using this approach:

 a. Fx = Fapp*cos(theta) = 86.6 N Fy = Fapp*sin(theta) = 50.0 N Fgrav = m*g = 100 N Fnorm = Fgrav - Fy = 70 N Ffrict = mu*Fnorm = 21 N Fnet = 65.6 N, right a = Fnet/m = 5.47 m/s/s, right b. Fx = Fapp*cos(theta) = 65.5 N Fy = Fapp*sin(theta) = 45.9 N Fgrav = m*g = 80 N Fnorm = Fgrav - Fy = 34.1 N Ffrict = mu*Fnorm = 6.8 N Fnet = 58.7 N, right a = Fnet/m = 7.34 m/s/s, right

Several additional examples and discussion of these Fnet=m*a problems are provided elsewhere at this site.

Net Force Problems Revisited (17 seconds)

38. The following diagrams depict an object of known mass upon an incline with a known coefficient of friction and angle. For each case, determine all missing quantities.

 a. m = 12 kg Fparallel = _______ Fperpendicular = _______ Fnet = __________, ________ (dir'n) a = __________, ________ (dir'n) b. m = 8 kg Fparallel = _______ Fperpendicular = _______ Fnet = __________, ________ (dir'n) a = __________, ________ (dir'n)

This inclined plane problem is another sure bet for the upcoming test (once more the numbers would be different). Such problems could involve frictionless situations (which are far easier than the case here) or surfaces where friction must be accounted for (similar the problem above). Such problems can be instantly simplified by resolving the force of gravity into parallel and perpendicular components using the equations:

 Fparallel = m*g* sin(theta) Fperpendicular = m*g* sin(theta)

where theta is the incline angle (given in each case) and g = 10 m/s/s (approximately). Once resolved, the Fgrav no longer needs to be considered since it has been replaced or substituted for by its components. The value of this strategy is that all forces are now in opposite directions. The Fperpendicular and Fnorm will balance each other (as is always the case on inclined planes) and Fparallel is opposed by Ffrict. The friction force is found using the equation

Ffrict = mu*Fnorm

where Fnorm is the same value as Fperpendicular and mu is the coefficient of friction (given in each case). Now with all forces known, the net force can be found by subtracting Ffrict from Fparallel. The acceleration is then determined using the equation

Fnet = m*a

This method will yield the following answers.

 a. Fgrav = m*g = 120 N Fparallel = m*g* sin(theta) = 31.1 N Fperpendicular = m*g* sin(theta) = 115. 9 N Fnorm = 115.9 N Ffrict = (0.1)*(115.9 N) = 11.6 N Fnet = 31.1 N - 11.6 N = 19.5 N a = (19.5 N)/(12 kg) = 1.62 m/s/s b. Fgrav = m*g = 80 N Fparallel = m*g* sin(theta) = 27.4 N Fperpendicular = m*g* sin(theta) = 75.2 N Fnorm = 75.2 N Ffrict = (0.15)*(75.2 N) = 11.3 N Fnet = 27.4 N - 11.3 N = 16.1 N a = (16.1 N)/(8 kg) = 2.01 m/s/s

More examples and further discussion of inclined plane problems can be found elsewhere at this site.

Inclined Planes (25 seconds)