Answer:
The net torque on the square plate is 2.72 N-m.
Explanation:
Given that,
Side = 0.2 m
Force [tex]F_{1}=18\ N[/tex]
Force [tex]F_{2}=26\ N[/tex]
Force [tex]F_{3}=14\ N[/tex]
We need to calculate the torque due to force F₁
Using formula of torque
[tex]\tau_{1}=-F_{1}d_{1}[/tex]
[tex]\tau_{1}=-F_{1}\times\dfrac{a}{2}[/tex]
Put the value into the formula
[tex]\tau_{1}=-18\times\dfrac{0.2}{2}[/tex]
[tex]\tau_{1}=-1.8\ N-m[/tex]
We need to calculate the torque due to force F₂
Using formula of torque
[tex]\tau_{2}=F_{2}d_{2}[/tex]
[tex]\tau_{2}=F_{2}\times\dfrac{a}{2}[/tex]
Put the value into the formula
[tex]\tau_{2}=26\times\dfrac{0.2}{2}[/tex]
[tex]\tau_{2}=2.6\ N-m[/tex]
We need to calculate the torque due to force F₃
Using formula of torque
[tex]\tau_{3}=F_{3}d_{3}[/tex]
[tex]\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}[/tex]
Put the value into the formula
[tex]\tau_{3}=0.1(14\sin45+14\cos45)[/tex]
[tex]\tau_{3}=1.92\ N-m[/tex]
We need to calculate the net torque on the square plate
[tex]\tau=\tau_{1}+\tau_{2}+\tau_{3}[/tex]
[tex]\tau=-1.8+2.6+1.92[/tex]
[tex]\tau=2.72\ N-m[/tex]
Hence, The net torque on the square plate is 2.72 N-m.
