Question

1. compute the sum of the torques about point a.

∑ta = 17 n·m
∑t=|f|×ℓ

Explanation:

Step1: Determine the torque formula

Torque $\tau = r\times F$, where $r$ is the lever - arm (perpendicular distance from the axis of rotation to the line of action of the force) and $F$ is the force. The sign of the torque: counter - clockwise is positive and clockwise is negative.

Step2: Calculate the torque due to each force

• For the $- 30N$ force: The lever - arm $r_1=7m$. The torque $\tau_1=-30\times7=-210N\cdot m$ (clockwise).
• For the $80N$ force: The lever - arm $r_2 = 2m$. The torque $\tau_2=80\times2 = 160N\cdot m$ (counter - clockwise).
• For the $40N$ force: The lever - arm $r_3=3m$. The torque $\tau_3=40\times3=120N\cdot m$ (counter - clockwise).

Step3: Calculate the sum of torques

$\sum\tau=\tau_1+\tau_2+\tau_3$.
Substitute the values: $\sum\tau=-210 + 160+120$.

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Answer:

$70N\cdot m$