Question

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with

Answer 1

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10\times (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r\times T$

$\tau=0.17\times 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.