Question

# A 12.3 kg box sits on a horizontal table. A string with tension 31.1 N pulls...

A 12.3 kg box sits on a horizontal table. A string with tension 31.1 N pulls on the box to the right, but static friction between the box and the table prevents the box from moving. The coefficient of static friction between the box and the table is 0.43. What is the magnitude of the static friction force on the box?

Maximum value of static friction is given by: N, where is coefficient of friction and N is normal reaction.

In the given problem, since there is no vertical acceleration, vertical forces must balance

=> Normal reaction N=weight=mg, where m is mass and g is gravitational acceleration.

So, maximum value of static friction=N=mg

Here,=0.43,m=12.3 kg. So,maximum value of static friction can be:0.43*12.3*9.8= 51.83 N.

Now,force with which object is pulled towards right=31.1 N.(which is less than maximum value of satic friction)

So,the static friction can balance this force by adjusting its value to 31.1 N.

So,magnitude of static friction=31.1 N

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