A 4.28 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=11.6N at an angle theta=26.5degrees above the horizontal, as shown.

Question

What is the magnitude of the normal force acting on the block when the force F is acting on it?

If, instead, the floor has a coefficient of kinetic friction µ_k = 0.04, what is the magnitude of the frictional force on the block when the block is moving?

What is the magnitude of the acceleration of the block when friction is being considered?

Answer 1

Solution:

1- The magnitude of the normal force is F_N = Sum of vertical forces on the block

= Weight of block - vertical component of force = W - F sinθ = mg - F sinθ

F_N = 4.28 kg ( 9.81 m/s²) - 11.6 N sin 26.5° = 36.8 N

2- The friction force F_f = μ_k F_N = 0.04 ( 36.8 N) = 1.47 N

3- The acceleration of the block can be found from ∑F = ma , ∑F is sum of horizontal forces parallel to the plane of motion assuming positive to the right , F cos 26.5° - F_f = m a ,

11.6 cos26.5° - 1.47N = 4.28 kg( a) , 10.38N - 1.47N = 4.28 kg (a), 8.91 N = 4.28 kg( a)

a = 8.91 N / 4.28 kg = 2.08 m/s² to the right →