Question

Calculate the resultant torque from the following diagram

Answer 1

torque=force ×distance ( perpendicular from c.o.m.)

so ,,30N ,50N try to move clockwise

80N ,20N try to move anticlockwise

200N cause no torque because its passes through c o m (centre of mass )

so

net torque=(30×4 +50×5) -(80×3 +20×1)

=370 - 260

=110N

Answer 2

Answer: The resultant torque is 110 N-m.

Explanation:

Given that,

Force $F_{A}= 20 N$

Force $F_{B}= 30 N$

Force $F_{C}= 80 N$

Force $F_{D}= 200 N$

Force $F_{E}= 50 N$

Distance OB = 4 m

Distance OC = 3 m

Distance OA = 1 m

Distance OE = 5 m

We know that,

The torque is the product of the force and distance.

According to figure,

The torque at D is zero because this force passing from center of mass.

Force B, and force E are clockwise.

So, The torque is

$\tau_{B} = 30\ N\times4m$

$\tau_{B} = 120\ N-m$

The torque is

$\tau_{E} = 50\ N\times5m$

$\tau_{E} = 250\ N-m$

Force A and force C are anticlockwise.

So, The torque is

$\tau_{A} = 20\ N\times1m$

$\tau_{A} = 20\ N-m$

The torque is

$\tau_{C} = 80\ N\times3m$

$\tau_{C} = 240\ N-m$

Now, the resultant torque is

Resultant torque = clockwise torque - anticlockwise torque

$\tau = \tau_{B}+\tau_{E}-\tau_{A}+\tau_{C}$

$\tau=120+250-240+20$

$\tau= 110\ N-m$

Hence, The resultant torque is 110 N-m.