Video Transcript
A body 𝐴 of mass 240 grams rests
on a rough plane inclined to the horizontal at an angle whose sine is
three-fifths. It is connected by a light
inextensible string passing over a smooth pulley fixed to the top of the plane to
another body 𝐵 of mass 300 grams. If the system was released from
rest and body 𝐵 descended 196 centimeters in three seconds, find the coefficient of
friction between the body and the plane. Take 𝑔 equal to 9.8 meters per
square second.
We begin by sketching the
system. We are told that the sin of angle
𝛼 is equal to three-fifths. Using our knowledge of right angle
trigonometry and the Pythagorean triple three, four, five, we know that cosine of
𝛼, or cos of 𝛼, is equal to four-fifths. The mass of the two bodies is given
in grams. We know that there are 1000 grams
in one kilogram. This means that 240 grams is equal
to 0.24 kilograms. We divide our value in grams by
1000.
Body 𝐴 will, therefore, have a
force acting vertically downwards equal to 0.24𝑔, where 𝑔 is equal to 9.8 meters
per square second. Body 𝐵 has a mass of 300 grams,
and this is equal to 0.3 kilograms. Therefore, this body has a downward
force of 0.3 multiplied by 𝑔.
We have a light inextensible string
passing over a smooth pulley. This means that the tension
throughout the string will be equal. It also means that when released,
the system will travel with uniform acceleration. Body 𝐴 has a normal reaction force
perpendicular to the plane. As the plane itself is rough, there
will be a frictional force acting down the plane.
We will now use Newton’s second
law, which states that the sum of the net forces is equal to the mass multiplied by
the acceleration, to resolve parallel and perpendicular to the plane for body 𝐴 and
vertically for body 𝐵. The weight of body 𝐴 is acting
vertically downwards. Therefore, we need to find the
components of this that are parallel and perpendicular to the plane. Once again, using our knowledge of
right angle trigonometry gives us a force of 0.24𝑔 multiplied by cos 𝛼
perpendicular to the plane and 0.24𝑔 multiplied by sin 𝛼 parallel to the
plane.
There are three forces acting on 𝐴
parallel to the plane: the tension force, the frictional force, and this weight
component. This gives us the equation 𝑇 minus
0.24𝑔 multiplied by sin 𝛼 minus the frictional force 𝐹 𝑟 is equal to 0.24𝑎. Perpendicular to the plane, the sum
of the net forces is equal to 𝑅 minus 0.24𝑔 multiplied by cos 𝛼. The body is not moving in this
direction. Therefore, this is equal to
zero.
Rearranging the equation, we see
that the normal reaction force 𝑅 is equal to 0.24𝑔 multiplied by cos 𝛼. Finally, resolving vertically for
body 𝐵, where downwards is the positive direction, gives us 0.3𝑔 minus 𝑇 is equal
to 0.3𝑎. We can substitute in our values for
sin 𝛼 and cos 𝛼. This means that the normal reaction
force is equal to 1176 over 625. 0.24𝑔 multiplied by sin 𝛼 is
equal to 882 over 625.
Our next step is to calculate the
acceleration of the system, given the fact that body 𝐵 descended 196 centimeters in
three seconds. In order to do this, we will use
the equations of motion, also known as the SUVAT equations. The displacement of the body was
196 centimeters. This is equal to 1.96 meters, as
there are 100 centimeters in a meter. The body was released from rest, so
the initial velocity is zero meters per second. We are trying to calculate the
acceleration, and we are told the time is three seconds.
This means that we can use the
equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us
1.96 is equal to zero multiplied by three plus a half multiplied by 𝑎 multiplied by
three squared. The right-hand side simplifies to
4.5𝑎. Dividing both sides by 4.5 gives us
𝑎 is equal to 98 over 225. The acceleration of the system is
98 over 225 meters per square second. We can now substitute this value
into our equations.
We now have two unknowns left, the
tension 𝑇 and the frictional force 𝐹 𝑟. We know that the normal reaction
force is 1176 over 625 newtons. By adding 𝑇 and subtracting 49
over 375 to both sides of the bottom equation, we can calculate the tension force
𝑇. This gives us a value of 𝑇 equal
to 2107 over 750 newtons. We can substitute this into the top
equation. Rearranging this equation gives us
a frictional force 𝐹 𝑟 equal to 1617 over 1250 newtons.
We now need to calculate the
coefficient of friction. And we know that the frictional
force is equal to this coefficient of friction multiplied by the normal reaction
force. This means that the coefficient of
friction 𝜇 is equal to 𝐹 𝑟 divided by 𝑅. Typing this into the calculator
gives us a coefficient of friction 𝜇 equal to eleven sixteenths.
