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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TopNewest272 N . – Jatin Yadav · 3 years, 11 months ago

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– Siddharth Iyer · 3 years, 11 months ago

I am sorry but the correct answer is 320 N. Could you tell me how you approached the problem???Log in to reply

Draw proper FBD. Let \(N_{1}\) and \(N_{2}\) be the normal reactions between rod and block and between ground block respectively. Let \(f_{1} \) and \(f_{2}\) be corresponding frictions.

At the verge of slipping , \(f_{1} = 0.4N_{1} , f_{2} = 0.4N_{2}\)

On the rod , we can balance torque apout the hinge.

Anticlockwise torque due to friction and weight = clockwise torque due to normal reaction.

\(\Rightarrow \frac{mglsin\theta}{2} + 0.4N_{1}lcos\theta = N_{1}lsin\theta \) ,

\(\Rightarrow N_{1} = 200N\)

Analysing FBD of block ,

\(N_{2} = W_{block} + N_{1} = 600N \)

\( F \geq f_{1} + f_{2} = 0.4(N_{1} + N_{2}) \) = \(0.4 \times 800 = \fbox{320N} \) – Jatin Yadav · 3 years, 11 months ago

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– Rohan Rao · 3 years, 11 months ago

But when I solved the torque equation, I got \(N_1\) as 300N. Can you check why?Log in to reply

– Jatin Yadav · 3 years, 11 months ago

Sorry , i can't check without your equations.Log in to reply

– Rohan Rao · 3 years, 11 months ago

No sorry never mind, I got it.Log in to reply

– Siddharth Iyer · 3 years, 11 months ago

Thanks!!! I figured my mistake too. i was not considering torque due to normal reaction.Log in to reply