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Unformatted text preview: Chapter 6 Force and Motion Solutions to Homework Problems: 13, 20, 21, 26, 29, 37, 41, 47, 49 HRW06.13 A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. If the coefficient of static friction is 0.50, what minimum force magnitude is required from the rope to start the crate moving? If , what is the magnitude of the initial acceleration of the crate? Solution
FN Ff r M T θ Mg Newton's Second Law: cos sin in order to start moving The maximum value for friction is The minimum force has to be equal to the maximum possible friction in order for the object to start moving: cos sin Solving the set of equations: sin cos cos sin N cos sin Newton's Second Law: cos sin The kinetic friction is: cos sin The tension is found in
cos sin cos cos sin sin sin N m/s 1 HRW06.20 A loaded penguin sled weighing 80 N rests on a plane inclined at angle =20° to the horizontal. Between the sled and the plane , the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.15 What is the least magnitude of the force , parallel to the plane, that will prevent the sled from slipping down the plane? What is the minimum magnitude that will start the sled moving up the plane? What value of is required to move the sled up the plane at constant velocity? Solution If the sled is sliding down, the friction points upward and we need extra force upward to prevent the sled from moving
FN (mg)x θ mg F Ff r (mg)y Newton's Second Law: cos sin sin cos sin cos N cos sin If the force tries to pull the sled upwards, the friction changes its direction and now it points downward. It is still static until the sled becomes moving. Thus, the force must overcome the maximum static friction in order for the sled to start moving.
FN Ffr (m g)x θ mg (m g)y F Newton's Second Law: cos sin sin cos sin cos !N cos sin The sled is already moving, thus the friction is kinetic. The equations are the same as in , only the frictional force is calculated as 2 sin cos "N HRW06.21 Block # weighs 711 N. The coefficient of static friction between block and table is 0.25; angle is 30°, assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. B θ A Solution
FN Ff r B T3 T2 mB g A T3 T1 θ T2 mAg For block B, Newton's Second law:
# # # $ For block A, Newton's Second law:
% % For the knot, Newton's Second law: cos sin Replacing cos sin and
% with the expressions from the equations for the blocks: 3 The maximum weight can be supported when the friction is at maximum,
# cos sin
% # %
% sin # cos
% # cot # tan 1.03 N HRW06.25 Two blocks, of weights 3.6 N and 7.2 N, are connected by a massless string and slide down a 30° inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find the magnitude of the acceleration of the blocks and the tension in the string. Solution 3.5 m/s 0.21 N HRW06.26 Two blocks are connected over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. Angle of the incline is 30°. Block A slides down the incline at constant speed. What is the mass of block B?
A B θ Solution
FN mA mAgsinθ m Ag T T Ffr mAgcos θ mBg θ Newton's Second Law for block A: % sin cos % Newton's Second Law for block B: % % cos 4 # # # Solving the set of equations: # % cos % sin cos % sin # % sin cos kg HRW06.29 The two blocks ( kg& kg' are not attached to each other. The coefficient of static friction between the blocks is , but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block?
F m M Solution
F F21 Ff r m M F12 Ffr mg Mg N2 Newton's Second Law for block Newton's Second Law for block Both blocks travel with same acceleration, . . This force acts as a normal force between the blocks. The friction force is determined as: $ $ Minimum force is required when the friction force is maximum, The acceleration of the system is: The force between the blocks is: 5 The minimum force required is: !" N HRW06.37 What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29 km/h and the between tires and track is 0.32? Solution In order not to slide, the friction force must be able to provide the necessary force for a circular motion. Newton's Second Law for the circular motion is:
( $ $ The larger the friction, the smaller the radius can be. The smallest radius corresponds to the largest possible value of the friction,
( ( ( " m/s
m s m HRW06.41 A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley? Solution Newton's Second Law at the top of the hill
) ( ) ( ( * N Newton's Second Law at the bottom of the hill
( ( HRW06.47 An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted at angle =40° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface. 6 θ Solution
Flift,y θ Fli ft Flift,x mg θ Newton's Second Law + )sin , cos +)
+) cos ( cos ( sin
( tan ( tan m cm on a kg by a cord HRW06.49 A puck of mass kg slides in a circle of radius frictionless table while attached to a hanging cylinder of mass through a hole in the table. What speed keeps the cylinder at rest?
M R v m Solution 7 M T R v T m mg Newton's Second Law circling puck: hanging cylinder:
( - , ( - ( - ( - m/s 8 ...
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