### Video Transcript

A box of mass 33 grams was resting on a smooth horizontal table that has a smooth pulley fixed at each end. A light inextensible string passed over one of the pulleys, π sub π΄, and connected the box to body π΄ of mass 26 grams which was hanging freely below the pulley. Another similar string passed over pulley π sub π΅ and connected the box with body π΅ of mass 24 grams hanging freely vertically below this pulley. The system was released from rest. Find the force exerted on both of the pulleys, π sub π΄ and π sub π΅, rounding your answer to the nearest two decimal places. Take π equal to 9.8 meters per second squared.

We will begin by sketching a diagram to model this situation. We have a box of mass 33 grams resting on a smooth horizontal table. This means that there will be no frictional force between the box and the table. We are told that there was a smooth pulley fixed at each end. We have two bodies of mass 26 grams and 24 grams hanging freely below these pulleys, and these are attached by two light inextensible strings. This means that the strings have no mass under a fixed length. As the string is inextensible, we also know that the acceleration of the system will be equal throughout. Since the pulleys are smooth, the tensions in each of the strings will be equal.

In the left-hand string, we will let the tension be π sub π΄ and in the right-hand string, π sub π΅. By calculating these values of π sub π΄ and π sub π΅, weβll be able to find the force exerted on both of the pulleys. The three bodies of mass 33 grams, 26 grams, and 24 grams will have a weight force acting vertically downwards. This will be equal to the mass multiplied by gravity. We are told that π is equal to 9.8 meters per second squared.

However, as the masses are given in grams, we will use the units of centimeters per second squared. π is therefore equal to 980. 26 multiplied by 980 is equal to 25,480. This means that the 26-gram body has a downward force of 25,480 dynes. In the same way, the 24-gram body has a downward force of 23,520 dynes. The 33-gram box has a downward force of 32,340 dynes. And this is equal to the normal reaction force acting vertically upwards.

When the system is released from rest, the 26-gram body will accelerate downwards, the 24-gram body will accelerate upwards, and the 33-gram box will accelerate to the left. As already mentioned, as the string is inextensible, this value of π will be the same throughout. The magnitude of the acceleration will be the same for the whole system. Using Newtonβs second law, force is equal to mass multiplied by acceleration, we will now set up three equations in three unknowns to help us calculate π sub π΄ and π sub π΅.

After clearing some space, we begin by considering the 26-gram body. As this body is accelerating downwards, we will take this to be the positive direction, and the sum of its forces is equal to 25,480 minus π sub π΄. This must be equal to the mass multiplied by the acceleration, in this case, 26π. We can rearrange this equation such that π sub π΄ is equal to 25,480 minus 26π. We will call this equation one. Letβs now focus on the 24-gram body. Since this body is accelerating upwards, we will consider this to be the positive direction. This means that the sum of the forces is equal to π sub π΅ minus 23,520. Once again, this is equal to the mass multiplied by the acceleration, which is 24π. Simplifying this equation, we get π sub π΅ is equal to 23,520 plus 24π. We will call this equation two.

As already mentioned, when considering the 33-gram box, we know that the normal reaction force π
is equal to 32,340 dynes. Resolving horizontally where the positive direction is to the left, the direction the body is accelerating, the sum of the forces is equal to π sub π΄ minus π sub π΅, and this is equal to 33π.

We can now substitute the expressions for π sub π΄ and π sub π΅ in equations one and two into this equation. The left-hand side becomes 25,480 minus 26π minus 23,520 plus 24π. And this must be equal to 33π. 25,480 minus 23,520 is equal to 1,960, and negative 26π minus 24π is negative 50π. The equation simplifies to 1,960 minus 50π is equal to 33π. We can add 50π to both sides such that 83π is equal to 1,960. Dividing through by 83, we have π is equal to 1,960 divided by 83. The acceleration of the system has magnitude 1,960 over 83 centimeters per second squared.

We can now substitute this value back in to equation one and equation two. Clearing some space once again, we have π sub π΄ is equal to 25,480 minus 26 multiplied by 1,960 over 83. This is equal to 24,866.0241. We can repeat this process for π sub π΅, which is equal to 23,520 plus 24 multiplied by 1,960 over 83. Typing this into our calculator gives us 24,086.74699. Both of these tensions are measured in dynes. Since the tension forces acting at the pulleys are at right angles to one another, we can use a force triangle and the Pythagorean theorem to calculate the magnitude of the force acting on the pulley itself. Considering pulley π΄, we have the two tension forces acting on the pulley as shown. The resultant force π sub π΄ will bisect these two forces. We can then create a right triangle as shown.

The Pythagorean theorem states that π squared plus π squared is equal to π squared, where π is the length of the longest side known as the hypotenuse. π sub π΄ squared is therefore equal to π sub π΄ squared plus π sub π΄ squared. The right-hand side simplifies to two π sub π΄ squared. We can then square root both sides of this equation so that π sub π΄ is equal to the square root of two π sub π΄ squared. We can now substitute in the value of π sub π΄ to calculate π sub π΄. π sub π΄ is equal to 35,165.86852.

We were asked to round this to two decimal places. The force exerted on pulley π sub π΄ is therefore equal to 35,165.87 dynes. We can repeat this process for π sub π΅ where this is equal to the square root of two multiplied by π sub π΅ squared. Substituting in our value of π sub π΅ and rounding to two decimal places gives us 34,063.80. The force acting on π sub π΅ is 34,063.80 dynes. We have now calculated the forces exerted on both pulleys.