Answer
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Hint: We can calculate the resultant torque directly by using the equation for calculating the torque. For each force given in the question, we need to find the torque and finally add them up to get the resultant value of torque.
Complete step by step solution:
We know that torque is defined as the product of force and perpendicular distance from the centre of mass. Mathematically, $\tau = F \times r$
Now, we need to calculate the value of torque at each point where force is acting. Clearly, the force at point A and C are tending to move anticlockwise, so we need to take their direction as negative. Similarly, the force at points B and D are tending to move the body clockwise, so we need to take their direction as positive.
Force at point ‘A’ is given$ - 20N$and perpendicular distance, $r = 3m$
Torque at point $A,{\tau _A} = - 20 \times 3 = - 60Nm$
Force at point ‘B’ is given $30N$ and perpendicular distance, $r = 4m$
Torque at point $B,{\tau _B} = 30 \times 4 = 120Nm$
Force at point ‘C’ is given$ - 80N$and perpendicular distance, $r = 3m$
Torque at point $C,{\tau _C} = - 80 \times 3 = - 240Nm$
Force at point ‘D’ is given$50N$and perpendicular distance, $r = 5m$
Torque at point $D,{\tau _D} = 50 \times 5 = 250Nm$
Now, for finding the net torque, we need to add the values of all the torque.
Therefore, we can write the net torque as,
$\tau = {\tau _A} + {\tau _B} + {\tau _C} + {\tau _D}$
$ \Rightarrow \tau = - 60 + 120 - 240 + 250$
$\therefore \tau = 360 - 300 = 60Nm$
Hence, the value of the required torque is $60Nm$.
Note: We should not get confused while taking the sign of the force on the body. One should not take positive signs for force acting on the body and negative signs for the force acting away from the body. While taking the sign of the forces, one should always consider the point of motion and accordingly should take the sign for clockwise and anticlockwise motion.
Complete step by step solution:
We know that torque is defined as the product of force and perpendicular distance from the centre of mass. Mathematically, $\tau = F \times r$
Now, we need to calculate the value of torque at each point where force is acting. Clearly, the force at point A and C are tending to move anticlockwise, so we need to take their direction as negative. Similarly, the force at points B and D are tending to move the body clockwise, so we need to take their direction as positive.
Force at point ‘A’ is given$ - 20N$and perpendicular distance, $r = 3m$
Torque at point $A,{\tau _A} = - 20 \times 3 = - 60Nm$
Force at point ‘B’ is given $30N$ and perpendicular distance, $r = 4m$
Torque at point $B,{\tau _B} = 30 \times 4 = 120Nm$
Force at point ‘C’ is given$ - 80N$and perpendicular distance, $r = 3m$
Torque at point $C,{\tau _C} = - 80 \times 3 = - 240Nm$
Force at point ‘D’ is given$50N$and perpendicular distance, $r = 5m$
Torque at point $D,{\tau _D} = 50 \times 5 = 250Nm$
Now, for finding the net torque, we need to add the values of all the torque.
Therefore, we can write the net torque as,
$\tau = {\tau _A} + {\tau _B} + {\tau _C} + {\tau _D}$
$ \Rightarrow \tau = - 60 + 120 - 240 + 250$
$\therefore \tau = 360 - 300 = 60Nm$
Hence, the value of the required torque is $60Nm$.
Note: We should not get confused while taking the sign of the force on the body. One should not take positive signs for force acting on the body and negative signs for the force acting away from the body. While taking the sign of the forces, one should always consider the point of motion and accordingly should take the sign for clockwise and anticlockwise motion.
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