solve it quickly. Calculate the net torque on the system about the point o as shown in figure if F= 11 N, F2 = 9 N,F3 = 10 N, a = 10 cm and b= 20 cm. (All the forces along the tangent.)
Answers
We know that, moment of force (torque) T = F * r.
As it is a cross product, T = F.rsinф. T1 = F1 * r1 = 11 *(20) * 1 = 220 N.
T1 = F2 * r2 = 9 * (20) * √3/2 = 90√3 N. T3 = 10* (10) *1/2 = 50 N.
Total Torque T= T1+ T2+T3 = 220 + 90√3+ 50 = 90(3 + √3) N.
Answer:
Net torque on the system about the point O is 3 N-m clockwise.
Explanation:
Given,
F= 11 N, F2 = 9 N,F3 = 10 N, a = 10 cm or 0.1 m and b= 20 cm or 0.1m.
To find: The net torque on the system about the point O.
Solution:
All the forces are acting along tangent. So we need not to check angles.
Anticlockwise is taken postive and clockwise is taken negative.
Net torque will summation of all forces with their sign.
τ = - F₁ . b - F₂ .b + F₃.a
τ = - (F₁ + F₂) b + F₃a
τ = - (11 + 9) * 0.2 + 10 × 0.1
τ = -4 + 1 Nm
or, τ = - 3 Nm
or, τ = 3 N - m ( clockwise)
Therefore, Net torque on the system about the point O is 3 N-m clockwise.