Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The net torque [tex]\tau = 21.95N \cdot m[/tex]
Explanation:
From the question we are told that
The length of each side is [tex]L = 2.5 m[/tex]
The first force is [tex]F_1 = 18N[/tex]
The second force is [tex]F_2 = 20 N[/tex]
The third force is [tex]F_3 = 11N[/tex]
The free body diagram for the question is shown on the second uploaded image
Generally torque is mathematically represented as
[tex]\tau = r * F[/tex]
Where [tex]\tau[/tex] is the torque
r is the length from the rotating point to the point the force is applied, this is also the radius of the circular path made
F is the force causing the rotation.
looking at the free body diagram we can deduce that L is the diameter of the circular path made as a result of toque
Now for the torque due to force [tex]F_1[/tex]
[tex]\tau_1 = - F_1 * r_1[/tex]
The negative sign is because the direction of [tex]F_1[/tex] is clockwise
=> [tex]\tau_1 = - F_1 * \frac{L}{2}[/tex]
Substituting value
[tex]\tau_1 = - 18 * \frac{2.5}{2 }[/tex]
[tex]\tau_1 = - 22.5 N \cdot m[/tex]
The torque as a result of the second force is mathematically evaluated as
[tex]\tau_2 = F_2 * r_2[/tex]
[tex]\tau_2 = F_2 * \frac{L}{2}[/tex]
[tex]= 20 * \frac{2.5}{2}[/tex]
[tex]\tau_ 2 = 25 \ N \cdot m[/tex]
The torque as a result of the third force is mathematically evaluated as
[tex]\tau_3 =r_3 (F_3 sin \theta + F_3 cos \theta )[/tex]
[tex]\tau_3 = \frac{L}{2} (F_3 sin \theta + F_3 cos \theta )[/tex]
Where the free body diagram [tex]\theta = 45^o[/tex]
[tex]\tau_3 = \frac{2.5}{2} (11 * sin (45) +11 cos (45) )[/tex]
[tex]\tau_ 3 = 19.45 \ N \cdot m[/tex]
The net torque the mathematically
[tex]\tau = \tau_1 + \tau_2 + \tau_3[/tex]
substituting value
[tex]\tau = -22.5 + 25 + 19.45[/tex]
[tex]\tau = 21.95N \cdot m[/tex]
