Question

1. compute the sum of the torques about point a. σta = 18 n·m σt = |f|×ℓ

Explanation:

Step1: Determine the sign of each torque

Torque is calculated as \(\tau = rF\sin\theta\). For a force, if it causes a counter - clockwise rotation about the point, the torque is positive; if it causes a clockwise rotation, the torque is negative.
The \(40N\) force causes a counter - clockwise rotation about point \(A\), so its torque \(\tau_1\) is positive. The \(- 30N\) and \(-90N\) forces cause clockwise rotations about point \(A\), so their torques \(\tau_2\) and \(\tau_3\) are negative.

Step2: Calculate each torque

• For the \(40N\) force: The perpendicular distance \(r_1 = 9m\). Using the torque formula \(\tau_1=r_1F_1\), we have \(\tau_1=(9m)\times(40N)=360N\cdot m\)
• For the \(-30N\) force: The perpendicular distance \(r_2 = 0m\) (since the line of action of the force passes through point \(A\)), so \(\tau_2=(0m)\times(- 30N)=0N\cdot m\)
• For the \(-90N\) force: The perpendicular distance \(r_3 = 4m\). Using the torque formula \(\tau_3=r_3F_3\), we have \(\tau_3=(4m)\times(-90N)=- 360N\cdot m\)

Step3: Sum the torques

\(\sum\tau_A=\tau_1+\tau_2+\tau_3\)
Substitute the values: \(\sum\tau_A = 360N\cdot m+0N\cdot m-342N\cdot m\)

Answer:

\(18N\cdot m\)