Question

The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.550mg is then applied at upward angle θ = 24°.  

(a) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μ_s = 0.610 and μ_k = 0.505?  

(b) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μ_s = 0.395 and μ_k = 0.295?

 What is the vertical component and horizontal component of the applied force? What is the gravitational force? What is the vertical acceleration? From Newton's second law for vertical motion, what is the normal force? What is fs,max? How does that value compare with the horizontal component of the applied force?

Answer 1

(a) The magnitude of acceleration of the block across the floor which is given as :

from a free body diagram, we have

on the y-direction :     F_app sin $\theta$ + F_N = F_g

F_app sin 24⁰ + F_N = mg

F_N = mg - (0.55 mg) sin 24⁰

F_N = mg [1 - (0.55) sin 24⁰]                                                          { eq.1 }

on the x-direction :        F_app cos $\theta$ - F_f = F_x

F_app cos 24⁰ - _{$\mu$ k} F_N = m a                                                          { eq.2 }

(0.55 mg) cos 24⁰ - _{$\mu$ k} mg [1 - (0.55) sin 24⁰] = m a

using, _{$\mu$ k} = 0.505

(because, the block is initially stationary and required a force greater than the maximum force of kinetic friction)

a = (0.55 g) cos 24⁰ - (0.505) g [1 - (0.55) sin 24⁰]

a = (0.55) (9.8 m/s²) (0.9135) - (0.505) (9.8 m/s²) [1 - (0.55) (0.4067)]

a = (4.92 m/s²) - (3.84 m/s²)

a = 1.08 m/s²

(b) The magnitude of acceleration of the block across the floor which is given as :

F_app cos 24⁰ - _{$\mu$ s} F_N = m a

(0.55 mg) cos 24⁰ - _{$\mu$ s} mg [1 - (0.55) sin 24⁰] = m a

using, _{$\mu$ s} = 0.395

(because, the block is initially stationary and required a force greater than the maximum force of static friction)

a = (0.55 g) cos 24⁰ - (0.395) g [1 - (0.55) sin 24⁰]

a = (0.55) (9.8 m/s²) (0.9135) - (0.395) (9.8 m/s²) [1 - (0.55) (0.4067)]

a = (4.92 m/s²) - (3 m/s²)

a = 1.92 m/s²