Physics 2211 Quiz #2 Solutions Fall 2007 Unless ...
Physics 2211 Fall 2007
Quiz #2
Solutions
gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A block of mass m is being slid to the right along a horizontal ceiling by an applied force of magnitude A that makes an angle θ with the vertical, as illustrated. The coefficient of static friction between the block and the ceiling is µs , while the coefficient of kinetic friction is µk . What is the acceleration magnitude the block in terms of any or all of m, A, µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
As announced during the quiz, the acceleration may depend on θ. Use Newton’s Second Law. Sketch a Free Body Diagram. X
X
Fx = Ax − fk = max
⇒
Fy = Ay − n − w = may = 0
A sin θ − µk n = max ⇒
n = A cos θ − mg
Substituting this expression for the normal force, n, into the x-equation above, A sin θ − µk (A cos θ − mg) = max
⇒
ax =
A sin θ − µk (A cos θ − mg) m
Quiz #2 Solutions Page 1 of 5
II. (16 points) Zach, of mass m, is riding upward in an elevator when the floor partially breaks free and hangs at an angle θ below the horizontal! Fortunately, the safety system of the elevator detects the problem and brings the elevator to a stop with acceleration magnitude a. Also fortunately, the coefficient of static friction, µs , is just sufficient to prevent Zach from sliding out of the gap as it slows. What is Zach’s apparent weight as the elevator slows, in terms of any or all of m, θ, µs , a, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Use Newton’s Second Law. Sketch a Free Body Diagram. The x axis has been chosen in the direction of the acceleration, so a = ax . Remember that the apparent weight of an object is the supporting force on that object. In this case, it is the normal force on Zach. X
Fx = w−fsx −nx = max
⇒
mg−fs sin θ−n cos θ = ma
⇒
mg−µs n sin θ−n cos θ = ma
Solve for n: mg − ma = µs n sin θ + n cos θ
⇒
n=
m (g − a) µs sin θ + cos θ
1. (6 points) In the problem above, how is the frictional force preventing Zach from sliding out of the gap related to his apparent weight? . . . . . . . . . . . . . . . . . . . . . . . Since Zach’s apparent weight is the normal force on him, and since the frictional force is proportional to the normal force (fs,max = µs n), then The frictional force magnitude is proportional to his apparent weight.
Quiz #2 Solutions Page 2 of 5
III. (16 points) In an electricity experiment, a plastic ball of mass m is attached to a string of length L and given an electric charge. The string is tied to a charged plane, inclined at an angle φ to the horizontal. The plane exerts an electrical force on the ball, perpendicular to the plane, causing the ball to swing up until the string is horizontal and remain there. What is the tension in the string, in terms of any or all of m, L, φ, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Use Newton’s Second Law. Sketch a Free Body Diagram. X
X
Fx = FEx − T = max = 0
Fy = FEy − w = may = 0
⇒ ⇒
T = FE sin φ mg = FE cos φ
Dividing the x-equation by the y-equation, T FE sin φ = mg FE cos φ
⇒
T = mg tan φ
~ 2. (6 points) In the problem above, how is the electric force, F~E , related ¯ ¯ to the tension in the string, T , and ~ ¯ ¯ the weight of the ball, w? ~ (Hint: Remember the notation, FE = FE , etc.) . . . . . . . . . . . . . . . . . . . . . . . The weight, the electric force, and the tension are the only forces acting on the ball. Since the ball is in equilibrium, the net force (i.e., the vector sum of the forces) must be zero, F~E + T~ + w ~ = 0, or F~E = −T~ − w ~
Quiz #2 Solutions Page 3 of 5
3. (10 points) An object is dropped from a helicopter. What is true when it reaches terminal velocity? (On Earth, do NOT neglect drag!) . . . . . . . . . . . . . . . . . . . . . . . The terminal velocity is a constant velocity, so the acceleration of the object is zero, and The net force on the object is zero.
4. (10 points) Harry the window washer swings from his bosun’s chair year after year, as shown on the left. One day he is washing windows near a flag pole, and for a change, ties one end of his rope to the pole instead of his chair, as shown on the right. How does the tension in Harry’s rope with one end tied to the pole compare to that with both ends tied to the chair? (On Earth.) The tension with one end tied to the pole is . . . . . . . . . . . . . . . . . . . . . . . . . . The weight of Harry and his bosun’s chair must be supported by the rope. With both ends attached to the chair, each end of the rope supports half the weight. With only one end attached to the chair, the tension in the rope must balance the entire weight. The tension with one end tied to the pole is twice as much as with both ends tied to the chair.
Quiz #2 Solutions Page 4 of 5
5. (10 points) In an experiment with a block of wood on an inclined plane, the following observations are made: (1) If the block is placed on the inclined plane, it remains there at rest. (2) If the block is given a small push, it accelerates towards the bottom without any further pushing. What conclusion can be drawn from these observations? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The coefficient of static friction must be greater than the coefficient of kinetic friction, or the kinetic frictional force would bring the block to a stop immediately upon it beginning to slide. The coefficient of static friction is greater than the coefficient of kinetic friction.
6. (10 points) The truck can accelerate to the right with a maximum acceleration magnitude amax if the box is not to slide. The forces on the box are that of weight w, ~ that of static friction f~s , and a normal force ~n. Suppose the truck accelerates at amax /2, only half its maximum acceleration. How do the magnitudes of those forces on the box in that situation compare to those when it accelerates at its maximum? (On Earth.) When the truck has acceleration amax /2 . . . . . . . . . . . . . . . . . . . . . . . . . . The weight and normal force are unaffected by the acceleration of the truck. Since the frictional force is the only horizontal force on the box, it is proportional to the horizontal acceleration. w is the same, n is the same, and fs is less then when the acceleration is amax .
Quiz #2 Solutions Page 5 of 5
Quiz #2
Solutions
gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) A block of mass m is being slid to the right along a horizontal ceiling by an applied force of magnitude A that makes an angle θ with the vertical, as illustrated. The coefficient of static friction between the block and the ceiling is µs , while the coefficient of kinetic friction is µk . What is the acceleration magnitude the block in terms of any or all of m, A, µs , µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
As announced during the quiz, the acceleration may depend on θ. Use Newton’s Second Law. Sketch a Free Body Diagram. X
X
Fx = Ax − fk = max
⇒
Fy = Ay − n − w = may = 0
A sin θ − µk n = max ⇒
n = A cos θ − mg
Substituting this expression for the normal force, n, into the x-equation above, A sin θ − µk (A cos θ − mg) = max
⇒
ax =
A sin θ − µk (A cos θ − mg) m
Quiz #2 Solutions Page 1 of 5
II. (16 points) Zach, of mass m, is riding upward in an elevator when the floor partially breaks free and hangs at an angle θ below the horizontal! Fortunately, the safety system of the elevator detects the problem and brings the elevator to a stop with acceleration magnitude a. Also fortunately, the coefficient of static friction, µs , is just sufficient to prevent Zach from sliding out of the gap as it slows. What is Zach’s apparent weight as the elevator slows, in terms of any or all of m, θ, µs , a, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Use Newton’s Second Law. Sketch a Free Body Diagram. The x axis has been chosen in the direction of the acceleration, so a = ax . Remember that the apparent weight of an object is the supporting force on that object. In this case, it is the normal force on Zach. X
Fx = w−fsx −nx = max
⇒
mg−fs sin θ−n cos θ = ma
⇒
mg−µs n sin θ−n cos θ = ma
Solve for n: mg − ma = µs n sin θ + n cos θ
⇒
n=
m (g − a) µs sin θ + cos θ
1. (6 points) In the problem above, how is the frictional force preventing Zach from sliding out of the gap related to his apparent weight? . . . . . . . . . . . . . . . . . . . . . . . Since Zach’s apparent weight is the normal force on him, and since the frictional force is proportional to the normal force (fs,max = µs n), then The frictional force magnitude is proportional to his apparent weight.
Quiz #2 Solutions Page 2 of 5
III. (16 points) In an electricity experiment, a plastic ball of mass m is attached to a string of length L and given an electric charge. The string is tied to a charged plane, inclined at an angle φ to the horizontal. The plane exerts an electrical force on the ball, perpendicular to the plane, causing the ball to swing up until the string is horizontal and remain there. What is the tension in the string, in terms of any or all of m, L, φ, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Use Newton’s Second Law. Sketch a Free Body Diagram. X
X
Fx = FEx − T = max = 0
Fy = FEy − w = may = 0
⇒ ⇒
T = FE sin φ mg = FE cos φ
Dividing the x-equation by the y-equation, T FE sin φ = mg FE cos φ
⇒
T = mg tan φ
~ 2. (6 points) In the problem above, how is the electric force, F~E , related ¯ ¯ to the tension in the string, T , and ~ ¯ ¯ the weight of the ball, w? ~ (Hint: Remember the notation, FE = FE , etc.) . . . . . . . . . . . . . . . . . . . . . . . The weight, the electric force, and the tension are the only forces acting on the ball. Since the ball is in equilibrium, the net force (i.e., the vector sum of the forces) must be zero, F~E + T~ + w ~ = 0, or F~E = −T~ − w ~
Quiz #2 Solutions Page 3 of 5
3. (10 points) An object is dropped from a helicopter. What is true when it reaches terminal velocity? (On Earth, do NOT neglect drag!) . . . . . . . . . . . . . . . . . . . . . . . The terminal velocity is a constant velocity, so the acceleration of the object is zero, and The net force on the object is zero.
4. (10 points) Harry the window washer swings from his bosun’s chair year after year, as shown on the left. One day he is washing windows near a flag pole, and for a change, ties one end of his rope to the pole instead of his chair, as shown on the right. How does the tension in Harry’s rope with one end tied to the pole compare to that with both ends tied to the chair? (On Earth.) The tension with one end tied to the pole is . . . . . . . . . . . . . . . . . . . . . . . . . . The weight of Harry and his bosun’s chair must be supported by the rope. With both ends attached to the chair, each end of the rope supports half the weight. With only one end attached to the chair, the tension in the rope must balance the entire weight. The tension with one end tied to the pole is twice as much as with both ends tied to the chair.
Quiz #2 Solutions Page 4 of 5
5. (10 points) In an experiment with a block of wood on an inclined plane, the following observations are made: (1) If the block is placed on the inclined plane, it remains there at rest. (2) If the block is given a small push, it accelerates towards the bottom without any further pushing. What conclusion can be drawn from these observations? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The coefficient of static friction must be greater than the coefficient of kinetic friction, or the kinetic frictional force would bring the block to a stop immediately upon it beginning to slide. The coefficient of static friction is greater than the coefficient of kinetic friction.
6. (10 points) The truck can accelerate to the right with a maximum acceleration magnitude amax if the box is not to slide. The forces on the box are that of weight w, ~ that of static friction f~s , and a normal force ~n. Suppose the truck accelerates at amax /2, only half its maximum acceleration. How do the magnitudes of those forces on the box in that situation compare to those when it accelerates at its maximum? (On Earth.) When the truck has acceleration amax /2 . . . . . . . . . . . . . . . . . . . . . . . . . . The weight and normal force are unaffected by the acceleration of the truck. Since the frictional force is the only horizontal force on the box, it is proportional to the horizontal acceleration. w is the same, n is the same, and fs is less then when the acceleration is amax .
Quiz #2 Solutions Page 5 of 5
