Video Transcript
A body of mass 21 kilograms was
placed on a rough horizontal plane. A horizontal force of magnitude
144.06 newtons acted on the body, causing a uniform acceleration of 4.9 meters per
second squared. Determine the coefficient of
friction of the plane. Take 𝑔 equal to 9.8 meters per
second squared.
We will begin by drawing a
free-body diagram to represent the scenario in this question. We’re told that the body has a mass
of 21 kilograms. There will therefore be a downward
force of 21𝑔, where 𝑔 is the acceleration due to gravity equal to 9.8 meters per
second squared. Due to Newton’s third law, we will
have a normal reaction force acting in the opposite direction to this. This will act vertically
upwards. We are told that a horizontal force
of magnitude 144.06 newtons acts on the body.
As the horizontal plane is rough,
there will be a frictional force 𝐅 𝑟 acting in the opposite direction to the
motion. Finally, we are told that the body
moves with a uniform acceleration of 4.9 meters per second squared. We are asked to calculate the
coefficient of friction of the plane. This coefficient of friction 𝜇 is
equal to the frictional force 𝐅 𝑟 divided by the normal reaction force 𝑅. We will therefore begin by
calculating these values by resolving horizontally and vertically.
Newton’s second law states that the
sum of the forces is equal to mass multiplied by acceleration. In the horizontal direction, we
have the 144.06-newton force and the frictional force. If we take the direction of motion
to be positive, we have 144.06 minus 𝐅 𝑟. This is equal to the mass of 21
kilograms multiplied by the acceleration of 4.9 meters per second squared. The right-hand side of our equation
simplifies to 102.9. We can then add the frictional
force and subtract 102.9 from both sides of our equation such that 𝐅 𝑟 is equal to
41.16 The frictional force is 41.16 newtons.
We can calculate the value of the
normal reaction force 𝑅 by resolving vertically. If we take the positive direction
to be vertically upwards, the sum of our forces is 𝑅 minus 21𝑔. As the body is not moving in this
direction, this is equal to zero. Adding 21𝑔 to both sides of our
equation, we have 𝑅 is equal to 21𝑔. Multiplying 21 by 9.8 gives us a
normal reaction force of 205.8 newtons.
We can now use both of these values
to calculate the coefficient of friction 𝜇. It is equal to 41.16 divided by
205.8, which simplifies to one-fifth or 0.2. The coefficient of friction of the
plane is one-fifth.
