Question

A mass of `100kg` is resting on a rough inclined plane of angle `30^(@)` If the coefficient of friction is `1//sqrt3` find the greatest and the least forces that acting parallel to the plane in both cases just maintain the mass in equilibrium .

Answer 1

Correct Answer - `100 kg f, zero` .
Here, `m = 100kg. theta = 30^(@) , mu = (1)/(sqrt(3))`
Force acting on the block down the plane
`= mg sin theta = 100 g sin 30^(@) = 50 g N = 50 kg f`
Force of friction `= f = mu R = mu mg cos theta`
`= (1)/sqrt 3 mgcos30^(@) = (1)/sqrt3 100 g sqrt3/(2) = 50 g N`
`=50 kg f`
As the force of friction just balances the force acting on the block down the plane therefore mass will be in equilibrium The least force required is zero
When we tend to move the mass up the inclined plane force of friction acts down the plane Therefore greatest force required just to start upward motion (or to maintain the block in equilibrium ) `= mg sin theta + f`
`50 kg f + 50 kg f = 100 kg f`
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