A 120 N force, directed at an angle θ above a horizontal floor, is applied to a 28.0 kg chair sitting on the floor. If θ = 0°, what are (a) the horizontal component Fh of the applied force and (b) the magnitude FN of the normal force of the floor on the chair? If θ = 34.0°, what are (c) Fh and (d) FN? If θ = 65.0°, what are (e) Fh and (f) FN? Now assume that the coefficient of static friction between chair and floor is 0.410. What is the maximum force of static friction on the chair if θ is (g) 0°, (h) 34.0°, and (i) 65.0°?

See explanation

Explanation:

Given:-

- The applied force Fa = 120 N

- The angle made with Fa and horizontal = θ

- The mass of chair m = 28.0 kg

Find:-

a) θ = 0, the horizontal component Fh of the applied force and (b) the magnitude FN of the normal force of the floor on the chair?

If θ = 34.0°, what are (c) Fh and (d) FN?

If θ = 65.0°, what are (e) Fh and (f) FN?

Now assume that the coefficient of static friction between chair and floor is 0.410. What is the maximum force of static friction on the chair if θ is (g) 0°, (h) 34.0°, and (i) 65.0°?

Solution:-

- We will consider the applied force (Fa) and develop a general expression for Fh as function of θ. Using Trigonometry we have:

Fh = Fa*cos (θ)

- The normal contact force (N) can be expressed similarly by applying equilibrium conditions on the chair in vertical direction.

N - mg - Fa*sin (θ) = 0

N = m*g + Fa*sin (θ)

For, θ = 0:

Fh = Fa*cos (0) = Fa

Fh = 120 N

N = (28*9.81) + 120*0

N = 274.68 N

For, θ = 34.0:

Fh = Fa*cos (34) = (120) * (0.82903)

Fh = 99.5 N

N = (28*9.81) + 120*sin (34)

N = 341.8 N

For, θ = 65.0:

Fh = Fa*cos (65) = (120) * (0.42261)

Fh = 50.7 N

N = (28*9.81) + 120*sin (65)

N = 383.4 N

- The maximum frictional force (Ff) is given by the following expression:

Ff = N*us

Where, us = coefficient of static friction = 0.41

For, θ = 0:

Ff = (274.68) * (0.41)

Ff = 112.62 N

For, θ = 34:

Ff = (341.8) * (0.41)

Ff = 140.13 N

For, θ = 65:

Ff = (383.4) * (0.41)

Ff = 157.2 N