Question

The coefficients of friction between a block of mass 2.0 kg and a surface are μs = 0.60 and μk = 0.25. Assume the only other force acting on the block is that due to gravity. What is the magnitude of the frictional force (in newtons) on the block if the block is at rest and the surface is inclined at 19o? Use g = 9.79 m/s2

Answer 1

The block will start moving if \[\tan \theta >\mu\](If you want to know why, please tell me).
\[\tan 19=0.344\]\[<\mu(0.6)\]
So therefore the block will remain at rest, and the frictional force will be \[mg \sin \theta\].
If u do not understand, tell me i will give you a diagram

Answer 2

Can you explain... .. why ??
\[\tan \theta > \mu\]

Answer 3

& your diagram will help me too ;)

Answer 4

As you see in the diagram, a force \[mg \sin \theta\] tries to bring the block down, while the frictional force \[mumg \cos \theta\] acting upwards resists that. So, in order for the body to move downward, the downward force must triumph over the friction.
\[mg \sin \theta >mumg \cos \theta\]
Cancel m and g from both sides and we get
\[\tan \theta >\mu\].
But since \[\theta=19\], \[\tan \theta <\mu\]. The downward force is unable to withstand the friction, and the body will be at rest. When the body is at rest, the force on both sides shall be equal. So the friction is \[mg \sin \theta\] and not \[mumg \cos \theta\]