A classical mechanics problem by Kartik Sharma

What is the coefficient of friction between a floor and a box weighing one ton-force if a minimum force of 600 kgf is required to start the box moving?

0.7 0.6 0.75 0.67

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1 solution

Kartik Sharma
Feb 3, 2017

This is a tricky one.

When is minimum force required? Yes, it is not given that force is applied horizontally.

Therefore, we have to find at what angle force must be applied so that we get a minimum force.

F - force applied \text{F - force applied}

μ coefficient of friction \mu - \text{coefficient of friction}

N - normal force \text{N - normal force}

m - mass of the block \text{m - mass of the block}

α angle force makes with the horizontal \alpha - \text{angle force makes with the horizontal}

For no slipping,

F cos ( α ) μ N \displaystyle F\cos(\alpha) \le \mu N

F sin ( α ) + N = m g \displaystyle F\sin(\alpha) + N = mg

Therefore,

F m g c o s ( α ) + μ sin ( α ) \displaystyle F \le \frac{mg}{cos(\alpha) + \mu \sin(\alpha)}

F m g 1 + μ 2 \displaystyle F \le \frac{mg}{\sqrt{1 + \mu^2}}

Therefore, to just move, this is the minimum force.

Hence,

μ = F 2 W 2 1 = 0.75 \mu = \frac{F^2}{W^2} - 1 = 0.75

The minimum force should be μ m g μ 2 + 1 \dfrac{\mu mg}{\sqrt{\mu^{2} + 1}} .

Sumanth R Hegde - 4 years, 4 months ago

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