Answer:
[tex]\tau=19.21\ N-m[/tex]
Explanation:
It is given that,
Mass of bundle of shingles, m = 10 kg
Upward acceleration of the shingles, [tex]a=1.5\ m/s^2[/tex]
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 :
[tex]T-mg=ma[/tex]
[tex]T=m(g+a)[/tex]
[tex]T=10\times (9.8+1.5)[/tex]
T = 113 N
Let [tex]\tau[/tex] is the minimum torque that the motor must be able to provide. It is given by :
[tex]\tau=r\times T[/tex]
[tex]\tau=0.17\times 113[/tex]
[tex]\tau=19.21\ N-m[/tex]
So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.
To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.
We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).
The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.
[tex]T-m.g=m.a\\T = m.g+m.a = m(g+a) = 10 kg (9.8m/s^{2}+1.5m/s^{2} )=113 N[/tex]
where,
Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.
[tex]\tau = r \times T = 0.17m \times 113N = 19N.m[/tex]
where,
To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.
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