Statics
FRICTION
Problem 507
The 2225-N block shown in Fig. P-507 is in contact with 45° incline. The coefficient of static friction is 0.25. Compute the value of the horizontal force P necessary to (a) just start the block up the incline or (b) just prevent motion down the incline. (c) If P = 1780 N, what is the amount and direction of the friction force?
Solution 507
Part (a) – Force P to just start the block to move up the incline
The force P is pushing the block up the incline. The push is hard enough to utilize the maximum allowable friction causing an impending upward motion.
answer Part (b) – Force P to just prevent the block to slide down the incline
In this case, the force P is not pushing the block upward, it simply support the block not to slide downward. Therefore, the total force that prevents the block from sliding down the plane is the sum of the component of P parallel to the incline and the upward friction force.
answer Part (c) – Force P = 1780 N
From the result of Part (b), the maximum force to prevent downward motion is 1335 N which is less than 1780 N. The force P therefore is pushing the block up the incline, thus, friction will act downward as shown below.
answer
Problem 512
A homogeneous block of weight W rests upon the incline shown in Fig. P-512. If the coefficient of friction is 0.30, determine the greatest height h at which a force P parallel to the incline may be applied so that the block will slide up the incline without tipping over.
Solution 512
Sliding up the incline
Tipping over
answer
Problem 527
A homogeneous cylinder 3 m in diameter and weighing 30 kN is resting on two inclined planes as shown in Fig. P-527. If the angle of friction is 15° for all contact surfaces, compute the magnitude of the couple required to start the cylinder rotating counterclockwise.
Solution
Problem 507
The 2225-N block shown in Fig. P-507 is in contact with 45° incline. The coefficient of static friction is 0.25. Compute the value of the horizontal force P necessary to (a) just start the block up the incline or (b) just prevent motion down the incline. (c) If P = 1780 N, what is the amount and direction of the friction force?
Solution 507
Part (a) – Force P to just start the block to move up the incline
The force P is pushing the block up the incline. The push is hard enough to utilize the maximum allowable friction causing an impending upward motion.
answer Part (b) – Force P to just prevent the block to slide down the incline
In this case, the force P is not pushing the block upward, it simply support the block not to slide downward. Therefore, the total force that prevents the block from sliding down the plane is the sum of the component of P parallel to the incline and the upward friction force.
answer Part (c) – Force P = 1780 N
From the result of Part (b), the maximum force to prevent downward motion is 1335 N which is less than 1780 N. The force P therefore is pushing the block up the incline, thus, friction will act downward as shown below.
answer
Problem 512
A homogeneous block of weight W rests upon the incline shown in Fig. P-512. If the coefficient of friction is 0.30, determine the greatest height h at which a force P parallel to the incline may be applied so that the block will slide up the incline without tipping over.
Solution 512
Sliding up the incline
Tipping over
answer
Problem 527
A homogeneous cylinder 3 m in diameter and weighing 30 kN is resting on two inclined planes as shown in Fig. P-527. If the angle of friction is 15° for all contact surfaces, compute the magnitude of the couple required to start the cylinder rotating counterclockwise.
Solution