Question

Jake from state farm is pulling a 144 kg refrigerator along the horizontal ground by a rope. The rope forms an angle of 15.0◦ above the horizontal. The coefficient of static friction between the ground and the refrigerator is 0.650. The coefficient of kinetic friction between the ground and the refrigerator is 0.435.
(a) If the refrigerator is initially at rest, what force must jake apply to the rope to get the refrigerator to move?

(b) Once the refrigerator begins moving, jaker applies a force equal to 600 N to the rope. What is the acceleration of the refrigerator?

Answer 1

given
m = 144 kg
theta = 15 degrees
mue_s = 0.65
mue_k = 0.435

a) let F is the required force.
let N is normal force acting on the refregerator.
Fnety = 0

F*sin(theta) + N - m*g = 0

N = m*g - F*sin(theta)

now apply, Fnetx = 0

F*cos(theta) - mue_s*N = 0

F*cos(theta) - mue_s*(m*g - F*sin(theta)) = 0

F*(cos(theta) + mue_s*sin(theta)) = mue_s*m*g

==> F = mue_s*m*g/(cos(theta) + mue_s*sin(theta))

= 0.65*144*9.8/(cos(15) + 0.65*sin(15) )

= 809 N <<<<<<<<<---------Answer

b) Fnetx = F*cos(theta) - mue_k*N

m*a = F*cos(theta) - mue_k*(m*g - F*sin(theta))

a = (F*cos(theta) - mue_k*(m*g - F*sin(theta)) )/m

= (600*cos(15) - 0.435*(144*9.8 - 600*sin(15) ))/144

= 0.231 m/s^2 <<<<<<<<<---------Answer