A wheel of radius 20 cm forces applied to it as shown in the figure. The torque produced by the forces 4 N at A, 8 N at B, 6 N at C at D at angles indicated is
1. 5.4 N-m anticlockwise
2. 1.80 N-m clockwise
3. 2.0 M-m clockwise
4. 3.6 n-m clockwise
Explain the solution clearly
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ANSWER
Given: A wheel of radius 20 cm has forces applied to is as shown in the figure.
To find the torque produced by the forces 4 N at A, 8 N at B, 6 N at C and 9 N at D at angles indicated
Solution:
We know,
τ=
r
×
F
=r(Fsinθ)...........(i)
where τ is the torque, r is the radius and F is the force applied, θ is the angle at which the force is applied.
Given, r=20cm=0.2m
Now the torque produced by the forces 4 N at A is
τ=4N×sin(90
∘
)×0.2=0.8Nm (by using the eqn(i) and substituting corresponding values from the figure)
in the anticlockwise direction.
The torque produced by the forces 8 N at B is
τ=8N×sin30×0.2=1.6×
2
1
=0.8Nm (by using the eqn(i) and substituting corresponding values from the figure)
in the clockwise direction.
The torque produced by the forces 6 N at C is
τ=0 as force and radius vectors are in the same direction
The torque produced by the forces 9 N at D is
τ=9N×sin90×0.2=1.8Nm (by using the eqn(i) and substituting corresponding values from the figure)
in the clockwise direction.
So Total torque is
τ=0.8−0.8−1.8=−1.8
Hence 1.8 Nm torque is produced in clockwise direction (negative sign indicates this)
Answer:
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