Video Transcript
Two masses of five 𝑚 and two 𝑚
kilograms are connected to each other by a light in extensible string passing over a
smooth pulley. Given that the system was released
from rest and that the force exerted on the pulley during the motion was 238
newtons, determine the value of 𝑚. Take 𝑔 equal to 9.8 metres per
second squared.
Let’s begin by looking at three key
words and what they tell us about this problem. As the string is light, we can
ignore the weight of the string. As the string is in extensible, the
accelerations are equal. This means that both bodies in the
system will move with the same acceleration. We also know that the pulley is
smooth, so there will be no friction. We can now draw a diagram with all
the forces involved in this problem.
The two bodies have masses five 𝑚
and two 𝑚. This means that they will have
downward forces of five 𝑚𝑔 and two 𝑚𝑔, respectively, where 𝑔 is gravity — in
this case 9.8 metres per second squared. In the string, there will be
vertical tension forces moving upwards. Once the system is released from
rest, the five 𝑚 body will accelerate vertically downwards. The two 𝑚 body will accelerate
vertically upwards with the same acceleration.
We will now be able to use Newton’s
second law 𝐹 equals 𝑚𝑎 or force equals mass times acceleration to set up two
simultaneous equations. As the body on the left-hand side
is moving downwards, the five 𝑚𝑔 force is positive and the tension force 𝑇 is
negative. Five 𝑚𝑔 minus 𝑇 is equal to five
𝑚 multiplied by 𝑎. We will call this equation one. As the body on the right-hand side
is moving upwards, the tension force is positive and the two 𝑚𝑔 force is
negative. 𝑇 minus two 𝑚𝑔 is equal to two
𝑚 multiplied by 𝑎. We will call this equation two.
We’re also told in the question
that the total force exerted on the pulley is 238 newtons. This means that the two tensions
must sum to 238. As the tensions are equal, two 𝑇
is equal to 238. We can divide both sides of this
equation by two. 238 divided by two is equal to
119. Therefore, the tension on each side
of the pulley is 119 newtons. We can then substitute this value
into equations one and two. We can calculate the acceleration
of the system by adding equations one and two.
On the left-hand side, five 𝑚𝑔
minus two 𝑚𝑔 is equal to three 𝑚𝑔. Negative 119 plus 119 is equal to
zero. On the right-hand side, five 𝑚𝑎
plus two 𝑚𝑎 is equal to seven 𝑚𝑎. We can cancel an 𝑚 on both sides
of the equation. This gives us three 𝑔 is equal to
seven 𝑎. Dividing both sides by seven gives
us a value of 𝑎 of three 𝑔 divided by seven. We’re told to take 𝑔 equal to
9.8. Three multiplied by 9.8 divided by
seven is equal to 4.2. Therefore, the acceleration of the
system is 4.2 metres per second squared.
We can now substitute this value
for 𝑎 back into equation one or equation two to calculate 𝑚. In this case, we will substitute 𝑎
equals 4.2 into equation two. Two multiplied by 9.8 is equal to
19.6 and two multiplied by 4.2 is equal to 8.4. Our equation becomes 119 minus
19.6𝑚 equals 8.4𝑚. Adding 19.6𝑚 to both sides of the
equation gives us 119 is equal to 28𝑚. We can then divide both sides by
28. 119 divided by 28 is equal to
4.25.
The value of 𝑚 is, therefore, 4.25
kilograms. We could use this value to work out
both masses. Five multiplied by 4.25 is equal to
21.25 and two multiplied by 4.25 is 8.5. The two masses are 21.25 kilograms
and 8.5 kilograms.
