Question. A block having equilateral triangular cross-section of side a and mass m is placed on a rough inclined surface,so tat it remains in equilibrium . The torque of normal force acting on the block about its center of mass is ?????
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3 Answers
49the torque due to normal force in the anticlockwise direction will be equal to the clockwise torque due to friction....
friction acting=downward force along the incline plane =mgsinθ
the dist of base of eq. Λ from centre of mass=a/2√3
thus torque due to normal reaction=mgasinθ/2√3..........hope this matches the answer.....
well it must...
1but the normal force acts along the center of mass only..........so torque should be zero.
Then why it has some value?????
although ur answer is correct.
49yeah there is just the question i was hoping......i had a similar misconception......bt the reason is here as it goes........
suppose u r standing and a friend is constantly pushing u....if u stand straight u will surely fall.....but to avoid that u spread your legs and in doing so u shift your normal reaction so as to balance the torques........
similar is the case for blocks......when there is a tangential force acting on a non circular body, it tries not to roll and so quite amazingly and magically shifts the normal reaction so as to balance the torques..............oterwise imagine a rolling block.........[1][1][1][1][3][3][4][4][4]