A body weighing 25.5 newtons rests
on a rough horizontal plane. A horizontal force acts on the
body, causing it to be on the point of moving. Given that the coefficient of
friction between the body and the plane is three seventeenths, determine the
magnitude of the force.
We can begin by sketching a
diagram. We know that the body weighs 25.5
newtons. There will be a normal reaction
force going vertically upwards of 𝑅 newtons. We will call the horizontal force
that is applied to the body 𝐹. As the horizontal plane is rough,
there will be a frictional force 𝐹 r newtons acting in the opposite direction to
this. We are told that the body is on the
point of moving, which means it is in limiting equilibrium. This tells us that the sum of the
vertical and horizontal forces equals zero. Resolving vertically, the forces up
are equal to the forces down. This means that 𝑅 is equal to
25.5. The normal reaction force is equal
to 25.5 newtons. Resolving horizontally, the force
𝐹 is equal to the frictional force.
We are also told that the
coefficient of friction is equal to three seventeenths. In limiting equilibrium, the
frictional force is equal to the coefficient of friction multiplied by the reaction
force. We can therefore calculate the
frictional force by multiplying three seventeenths by 25.5. This is equal to 4.5 newtons. As the horizontal force acting on
the body is equal to this, we know that this is equal to 4.5 newtons.