Video Transcript
A body of mass 162 grams rests on a
rough plane inclined to the horizontal at an angle whose tangent is four-thirds. It is connected by a light
inextensible string passing over a smooth pulley fixed to the top of a plane to
another body of mass 181 grams hanging freely vertically below the pulley. The coefficient of friction between
the first body and the plane is one-half. Determine the distance covered by
the system in the first seven seconds of its movement, given that the bodies were
released from rest. Take 𝑔 equal to 9.8 meters per
square second.
We will begin by sketching the
system. We are told that the plane is
inclined to the horizontal at an angle 𝛼, where tan or the tangent of 𝛼 is equal
to four-thirds. Using our knowledge of right-angle
trigonometry, we know that tan 𝛼 is equal to the opposite over the adjacent. As we have a three-four-five
Pythagorean triple, we know that sin 𝛼 is equal to four-fifths and cos or cosine of
𝛼 is equal to three-fifths.
The two bodies have a mass of 162
grams and 181 grams. As there are 1000 grams in a
kilogram, these are equal to 0.162 kilograms and 0.181 kilograms, respectively. We are told that gravity is equal
to 9.8 meters per square second, which means that the downward vertical force of
body A will be equal to 0.162 multiplied by 𝑔. The vertical downward force of body
B will be equal to 0.181 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 both bodies will
move with the same acceleration. We are told that the plane is
rough, which means there will be a frictional force 𝐹 r acting against the
motion. In this case, as body A moves up
the slope, the frictional force will act down the slope. We have a normal reaction force
acting perpendicular to 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, more commonly written as 𝐹 equals 𝑚𝑎. For body B, we will resolve
vertically and for body A we will resolve parallel and perpendicular to the
plane.
In order to do this, we need to
find these two components of the weight force, 0.162 multiplied by 𝑔. Using our knowledge of right-angle
trigonometry, we see that the force perpendicular to the plane is equal to 0.162𝑔
multiplied by cos 𝛼. The component of the force parallel
to the plane is equal to 0.162𝑔 multiplied by sin 𝛼. We know that sin 𝛼 is equal to
four-fifths or 0.8 and cos 𝛼 is equal to three-fifths or 0.6.
For body A resolving perpendicular
to the plane, the sum of the net forces is equal to 𝑟 minus 0.162𝑔 multiplied by
0.6. This is equal to 0.162 multiplied
by zero as the body is not accelerating in this direction. We can simplify this by multiplying
0.162, 9.8, and 0.6. This gives us 𝑟 minus 0.95256 is
equal to zero. Adding 0.95256 to both sides gives
us a value of 𝑟 equal to this number. The normal reaction force equals
0.95256 newtons.
We can now resolve parallel to the
plane for body A. We have three forces acting in this direction: the tension in the
string, the frictional force, and 0.162𝑔 multiplied by 0.8. As the body is moving up the plane,
we have 𝑇 minus 𝐹 r minus 0.162𝑔 multiplied by 0.8 is equal to 0.162𝑎. This simplifies to 𝑇 minus 𝐹 r
minus 1.27008 is equal to 0.162𝑎.
We know that when dealing with a
rough surface, the frictional force 𝐹 r is equal to 𝜇, the coefficient of
friction, multiplied by the normal reaction force. In this question, we are told that
𝜇 is equal to one-half. The frictional force is therefore
equal to one-half of the normal reaction force. As this is equal to 0.95256, we can
multiply this by one-half and substitute it into our equation. This gives us 𝑇 minus 1.74636 is
equal to 0.162𝑎.
Let’s now consider body B and
resolve in a vertical direction. As the body is accelerating
downwards, we will take this to be the positive direction. We have 0.181𝑔 minus 𝑇 is equal
to 0.181𝑎. Multiplying 0.181 by 9.8 gives us
1.7738. We now have two equations with two
unknowns, the tension 𝑇 and acceleration 𝑎. We will clear some space and solve
these simultaneous equations by elimination. Adding equation one and two will
eliminate 𝑇. The left-hand side becomes 0.02744
and the right-hand side becomes 0.343𝑎. We can divide both sides by 0.343
so that 𝑎 is equal to 0.08. The acceleration of the system when
released from rest is 0.08 meters per square second.
We can now use this value to
calculate the distance covered by the system in the first seven seconds. We will do this using our equations
of motion or SUVAT equations. We need to calculate the
displacement 𝑠. We know that the initial velocity
is zero meters per second, the acceleration is 0.08 meters per second squared, and
the time taken is seven seconds. We will use the equation 𝑠 is
equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us
𝑠 is equal to zero multiplied by seven plus a half multiplied by 0.08 multiplied by
seven squared. This is equal to 1.96. The distance covered is therefore
equal to 1.96 meters. As there are 100 centimeters in one
meter, this is equal to 196 centimeters.
The system covers a distance of 196
centimeters during the first seven seconds.
