Question

A man is pulling a crate of mass m=50kg using a massless rope. The rope is at an angle of theta=30 degrees with respect to the floor. There is friction between the crate and the floor. The man pulls the rope with an increasing force until the block starts to move (let's call Fa the force he needs to apply to start moving the block). Once the block starts moving, the man then changes his pulling force (increasing or decreasing it) so that the block moves with constant velocity (let's call Fb the force that he needs to apply to keep the block moving with constant velocity). The coefficients of friction are given by u(s)=0.7 and u(k)=0.5. Write an equation and compute the final result for the value of Fa and the value of Fb.

Answer 1

a)
let m = 50 kg

theta = 30 degrees

let N1 is the normal force acting on the mass just before the mass start moving.

Apply, Fnety = 0

N + Fa*sin(30) - m*g = 0

N = m*g - Fa*sin(30)

now apply, Fnetx = 0

Fa*cos(30) - fs_max = 0

Fa*cos(30) - mue_s*N = 0

Fa*cos(30) - mue_s*(m*g - Fa*sin(30)) = 0

Fa*cos(30) - mue_s*m*g + mue_s*Fa*sin(30) = 0

Fa*(cos(30) + mue_s*sin(30) ) = mue_s*m*g

==> Fa = mue_s*m*g/(cos(30) + mue_s*sin(30) )

= 0.78*50*9.8/(cos(30) + 0.7*sin(30))

= 314 N <<<<<<<<<<------------------------Answer

b) simillarly,

Fb = mue_k*m*g/(cos(30) + mue_k*sin(30) )

= 0.5*50*9.8/(cos(30) + 0.5*sin(30))

= 210 N <<<<<<<<<<------------------------Answer