A uniform bar AB 10 m long and weighing 280 N is hinged at B and rests upon a 400 N block as shown. If the coefficient of friction at all contact surfaces is 0.4. The value of horizontal force F required to start the motion of the 400 N block is...?
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272 N .
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I am sorry but the correct answer is 320 N. Could you tell me how you approached the problem???
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Ok i got my mistake , i was not considering torque due to friction on the rod.
Draw proper FBD. Let N1 and N2 be the normal reactions between rod and block and between ground block respectively. Let f1 and f2 be corresponding frictions.
At the verge of slipping , f1=0.4N1,f2=0.4N2
On the rod , we can balance torque apout the hinge.
Anticlockwise torque due to friction and weight = clockwise torque due to normal reaction.
⇒2mglsinθ+0.4N1lcosθ=N1lsinθ ,
⇒N1=200N
Analysing FBD of block ,
N2=Wblock+N1=600N
F≥f1+f2=0.4(N1+N2) = 0.4×800=320N
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N1 as 300N. Can you check why?
But when I solved the torque equation, I gotLog in to reply
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A uniform bar AB 10 m long and weighing 280 N is hinged at B and rests upon a 400 N block as shown. If the coefficient of friction at all contact surfaces is 0.4. The value of horizontal force F required to start the motion of the 400 N block is...? what is the solution