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The Classroom door is of width 50 cm. If the Handle of the door is 20 cm from the edge and Force of 5N is applied on the handle. Calculate the torque?

I have attached what I think is the answer. But the answer on the site from where I took this question is 0.8 Nm. Here is the answer the site gives:-

Handle of the door is situated at 20 cm. Therefore line of action is 20/2 = 10 cm.

Length of the lever arm = d = 50 - 10 = 40 cm = 0.4 m

Force applied = 2N

Torque = F × d

= 2N × 0.4 m

= 0.8 Nm.

If this above answer is correct then please please tell me how this line of action and lever arm is calculated.

I have attached what I think is the answer. But the answer on the site from where I took this question is 0.8 Nm. Here is the answer the site gives:-

Handle of the door is situated at 20 cm. Therefore line of action is 20/2 = 10 cm.

Length of the lever arm = d = 50 - 10 = 40 cm = 0.4 m

Force applied = 2N

Torque = F × d

= 2N × 0.4 m

= 0.8 Nm.

If this above answer is correct then please please tell me how this line of action and lever arm is calculated.