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

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest272 N .

Log in to reply

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

Log in to reply

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 ,

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

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply