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The uniform 50kg pole ABC is balanced in...

The uniform `50kg` pole `ABC` is balanced in the vertical position. A `500N` horizontal force is suddenly applied at `B`. If the coefficient of kinetic friction between the pole and the ground is `0.3`, determine the initial acceleration of point `A`. (Take `g=10ms^(-2)`).

Text Solution

Verified by Experts

`N=mg=500`
`5000muN=ma_(x)rArra_(x)=(500-muN)/m`
`rArra_(x)=((500)(-0.3)(500))/(50)=7 ms^(-2)`
Now, let us calculate torque due to forces about the centre of mass or centre of gravity `G` of the rod, then `tau=(500)(1)-(0.3)(500)(3)`
`rArrtau=500-450=50 Nm`

As `alpha=(tau)/IrArralpha=50/(1/12(50)(6)^(2))=rarr alpha=1/3 rads^(-2)`
So, acceleration of point `A` is
`a_(A)=a_(x)-ralpha=7-(3)(1/3)=6 ms^(-2)` (to the right)
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