Question

A teapot is initially at rest on a horizontal tabletop, then one end of the table is lifted slightly. Does the normal force increase or decrease? Does the force of static friction increase or decrease?

Solution

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The free-body diagram below describes the teapot before the table is lifted and after the table has been lifted.

Before the table is lifted, the teapot was at rest and the net force acting on the teapot in yy direction was Zero\textbf{Zero}, which could be represented mathematically as follows

Fy=FNmg=0\sum F_{y}=F_{N}-mg=0

FN=mgF_{N}=mg

Now, let's see what happens to the value of FNF_{N} after the table has been lifted, from the free-body diagram at the right side of the graph below, we can see that the forces acting on the teapot in yy direction are FNF_{N} and mgcos(θ)mg \cos{(\theta)}, and since the teapot still at rest, then the net force acting on the teapot in yy direction is Zero\textbf{Zero}, which could be represented mathematically as follows

Fy=FNmgcos(θ)=0\sum F_{y}=F_{N}-mg\cos{(\theta)}=0

FN=mgcos(θ)F_{N}=mg\cos{(\theta)}

Notice that [cos(θ)<1[\cos{(\theta)} < 1 for θ0]\theta \neq 0], hence

mgcos(θ)<mgmg\cos{(\theta)} < mg

Which means that the normal force on the teapot will decrease after lifting the table.

The force of static friction between the teapot and the horizontal surface at any moment before the teapot starts moving is equal to the horizontal force acting on the teapot at that moment. So, the horizontal force before the table is lifted was Zero\textbf{Zero} and became [mgsin(θ)][mg \sin{(\theta)}] after the table has been lifted, meaning that the horizontal force increased after the table has been lifted which in turns means that the force of static friction has increased.

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