A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0° with the horizontal.

Question

A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0° with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80% of that for the static case

a.    What is the coefficient of static friction?

b.    What is the maximum force that can be applied upward along the plane on the rope and not move the block?

c.     With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed?

d.    If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction?

e.    Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?

Answer 1

Define a coordinate system such that positive x is up the ramp. I will use g= 9.81 m/s² for the calculations.

a) Sum the x and y force components: ΣF_x = F_f - mg sin 37 = 0, ΣF_y = F_N - mg cos 37 = 0. Normal force F_N then = 783 Newtons. In the given situation, the friction force is the maximum static friction F_f = μ_s F_N = 590 N from the x equation, so static coefficient μ_s = 590/783 = 0.754.

b) Now with applied force F, ΣF_x = F - F_f = 0, so F = 590 N.

c) We need the kinetic friction coefficient μ_k = 0.8 μ_s = 0.603, yielding a friction force μ_k F_N = 472 N. With F > 590 N, net force is no longer zero: ΣF_x = F - F_f = ma, ma > 590 - 472 = 118 N, so a > 1.18 m/s². For constant speed, a = 0 and F = 472 N.

d) Once the crate is moving down, kinetic friction applies. ΣF_x = μ_k F_N - mg sin 37 = ma, ma = 472 - 590 = -118 N, a = -1.18 m/s².

e) With previous net force of -118 N, a positive F = 118 N will prevent acceleration.

Answer 2

On drawing the free body diagram at initial point

(a)tan theta= coefficient of static friction = tan 37 = 0.573

the dynamic friction is 80% = 08*0.573 =0.458A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0° with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80% of that for the static case

a.    What is the coefficient of static friction?

b.    What is the maximum force that can be applied upward along the plane on the rope and not move the block?

F = 100g (0.753 + 0.601)

(c) Reduced force = 100g(0.458 +0.601) = F

Acceration = Reduced force/Mass = F/100 m/s²

(d) Down ward accelartion = gSin theta -µ gCos theta , Where µ = 0458

use theta = 37 to get acceleration

(e) F = mgSin theta + µm gCos theta use m= 100 , g= 9.8 and theta = 37 degree