# The coefficient of static friction between a 1 kg box placed on a 30 degree ramp and the ramp is 0.65If the mass of the box can be increased in increments of 1 kg, what does it have to be made for...

The coefficient of static friction between a 1 kg box placed on a 30 degree ramp and the ramp is 0.65

If the mass of the box can be increased in increments of 1 kg, what does it have to be made for it to start moving down the ramp?

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### 1 Answer

The box in the problem is placed on a ramp that makes an angle of 30 degrees with the horizontal. The coefficient of static friction of the ramp is 0.65.

Let the mass of the box be M. A force due to the gravitational attraction of the Earth acts in a direction vertically downwards. This can be broken into two components: one component is acting along the ramp in the downward direction and is given by M*g*sin 30 and the other is normal to the surface of the ramp given by M*g*cos 30.

The force of resistance due to friction is Fr = 0.65*M*g*cos 30

If the box has to slide down, the component of force along the ramp of M*g*sin 30 should be greater than Fr. We see that both the resistive force due to friction as well as the component of force along the ramp have the term M, which gets canceled.

**This implies that changing the mass of the box by adding more blocks of 1 kg into it cannot make it slide down the ramp.**

Also, sin 30 = 0.5 and 0.65*cos 30 = 0.5629

As 0.5 < 0.5629, the box does not slide down the slope.