Question

3) A constant pulling force of 150 N applied at an angle of 21.7° above the horizontal, accelerates a crate across a rough horizontal surface. The has mass 35.2 kg, and the acceleration is 0.571 m/s. a) What is the normal force acting on the crate? b) What is the friction force acting on the crate? c) What is the coefficient of kinetic friction between surface?

Answer 1

let F = 150 N
theta = 21.7 degrees
m = 35.2 kg
a = 0.571 m/s^2

a) let N is the normal force acting.

Apply, Fnety = 0

N + F*sin(21.7) - m*g = 0

N = m*g - F*sin(21.7)

= 35.2*9.8 - 150*sin(21.7)

= 290 N <<<<<<<<<<---------------Answer

b) Apply, Fnetx = F*cos(21.7) - fk

m*a = F*cos(21.7) - fk

==> fk = F*cos(21.7) - m*a

= 150*cos(21.7) - 35.2*0.571

= 119 N <<<<<<<<<<---------------Answer

c) we know, fk = mue_k*N

==> mue_k = fk/N

= 119/290

= 0.410 <<<<<<<<<<---------------Answer

Answer 2

SOLUTION :

a.

N = mg - P sin(21.7) = 35.2*9.8 - 150*sin(21.7) = 289.5 N (ANSWER)

b.

P cos(21.7)  - F = m a 

=> F = P cos(21.7) - m a = 150 cos(21.7) - 35.2*0.571 = 119.27 N (ANSWER)

c.

F = µk * N

=> µk = F/N = 119.27/289.5 = 0.41 (ANSWER).