Friction Force

Question

A 11.5 kg sled is pulled with a 37.0 N force at a 40 degree angle, across the ground where Mk=0.110. What is the force of friction on the sled? (Unit=N)

Answer 1

I assume the 40º angle is above the horizontal.

 

f = force of friction

 

F_n = normal force

 

f = μ_kF_n

 

T = applied force

 

Tsinθ + F_n = mg ⇒ F_n = mg - Tsinθ

 

f = μ_k(mg - Tsinθ) = (0.110)[(11.5 kg)(9.81 m/s²) - (37.0 N)sin(40º)] = 9.79 N