Question

You wish to buy a motor that will be used to lift a 30-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s^2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.18 m.

Determine the minimum torque that the motor must be able to provide.

Answer 1

Torque = 61.02Nm

Step-by-step explanation:

Formula for calculating torque = rFsin(theta)

r is the radius = 0.18m

F is the force used to lift the car

F = m(a+g) (Newton's second law)

F = 30×(1.5+9.8)

F = 30×11.3

F= 339N

since the shingles accelerated upward, the angle it makes with the ground will be 90°

theta = 90°

Using the formula above of finding torque,

Torque = 339×0.18sin90°

Torque = 339×0.18(1)

Torque = 61.02Nm

Answer 2

Given Information:

Mass = m = 30 kg

acceleration = α = 1.5 m/s²

Radius of pulley = r = 0.18 m

Required Information:

Minimum torque = τ = ?

Answer:

Minimum torque = τ = 61.07 N.m

Step-by-step explanation:

There are two forces acting on the bundle of shingles;

One is acting downward force due to the weight and the other force is acting upward to lift the bundle of shingles.

Therefore, the sum of forces on the bundle of shingles is given by,

∑F = mα

Fup - Fdown = mα

Fup - mg = mα

Fup = mα + mg

Fup = m(α + g)

Fup = 30(1.5 + 9.81)

Fup = 339.3 N

The required minimum torque that the motor must be able to provide is given by

τ = Fup*r

τ = 339.3*0.18

τ = 61.07 N.m