[0702]

A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0 degrees. Once it has been set into motion, the angle is reduced to 30.0 degrees to keep the block moving at constant speed. Calculate the maximum force of static friction and the force of kinetic friction for the surfaces in contact?

[jiten55]

NB Point to note is that in both cases the net force down the inclined plane is zero.

In the Static case, it is obvious.

In the second case, net force is zero because acceleration is zero.

In both cases, component of weight along the plane = friction.

Friction force = coeff. Of friction times Normal Reaction.


1. Suppose coefficient of static friction = f

Normal Reaction = R = 7.80 g cos 35

Static Friction = fR = 7.80 g f cos 35

Component of weight along the inclined plane = 7.80 g Sin 35

7.80 g f cos 35 = 7.80 g Sin 35

f = tan 35

Coeff. Of Static Friction = tan 35

Maximum Static Friction Force = 7.80 g f cos 35

= 7.80 g Sin 35

2. Use same method as above.

Coeff. Of Kinetic Friction = tan 30

Maximum Kinetic Friction Force = 7.80 g Sin 30

[sGt HarDKorE]

Sorry for reviving such an old question but I have come across this question in studying for physics and I have a question, what is the g standing for in this problem? Gravity?

[Unknown008]

Yes, sGt :)