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# Rotational Dynamics Problem!

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...?

For a better image(not much better): http://bit.ly/18yUu8D

Note by Siddharth Iyer
4 years, 5 months ago

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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

- 2 months ago

272 N .

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Ok i got my mistake , i was not considering torque due to friction on the rod.

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 ,

$$N_{2} = W_{block} + N_{1} = 600N$$

$$F \geq f_{1} + f_{2} = 0.4(N_{1} + N_{2})$$ = $$0.4 \times 800 = \fbox{320N}$$

- 4 years, 5 months ago

But when I solved the torque equation, I got $$N_1$$ as 300N. Can you check why?

- 4 years, 5 months ago

Sorry , i can't check without your equations.

- 4 years, 5 months ago

No sorry never mind, I got it.

- 4 years, 5 months ago

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

- 4 years, 5 months ago