Question Video: Analysis of the Motion of Two Bodies Connected by a String over a Pulley Where One Is Resting on an Inclined Plane and the Other Hanging Vertically | Nagwa Question Video: Analysis of the Motion of Two Bodies Connected by a String over a Pulley Where One Is Resting on an Inclined Plane and the Other Hanging Vertically | Nagwa

# Question Video: Analysis of the Motion of Two Bodies Connected by a String over a Pulley Where One Is Resting on an Inclined Plane and the Other Hanging Vertically Mathematics • Third Year of Secondary School

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A body of mass 2.4 kg rests on a smooth plane inclined at an angle of 30Β° to the horizontal. It is connected by a light inextensible string passing over a smooth pulley, fixed at the top of the plane, to another body of mass 1.6 kg hanging freely vertically below the pulley. When the system was released from rest, the two bodies were on the same horizontal level. Then 10 seconds later, the string broke. Determine the time taken for the first body to start moving in the opposite direction after the string broke. Take π = 9.8 m/sΒ².

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### Video Transcript

A body of mass 2.4 kilograms rests on a smooth plane inclined at an angle of 30 degrees to the horizontal. It is connected by a light inextensible string passing over a smooth pulley, fixed at the top of the plane, to another body of mass 1.6 kilograms hanging freely vertically below the pulley. When the system was released from rest, the two bodies were on the same horizontal level. Then 10 seconds later, the string broke. Determine the time taken for the first body to start moving in the opposite direction after the string broke. Take π equal to 9.8 meters per square second.

Letβs begin by sketching the initial situation. We have a smooth plane inclined at an angle of 30 degrees. The two bodies have masses of 2.4 kilograms and 1.6 kilograms. This means they will have a vertical downward force equal to 2.4π and 1.6π, respectively, where gravity is equal to 9.8 meters per square second. 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 from rest, the magnitude of acceleration will be the same for the whole system.

In order to start to solve the problem, weβll use Newtonβs second law, which states that the sum of the net forces is equal to the mass multiplied by the acceleration. For body A, we will resolve parallel to the plane. And for body B, we will resolve vertically. As the weight force of body A is acting vertically downwards, we need to use our knowledge of right-angle trigonometry to calculate the components of the force parallel and perpendicular to the plane. The component parallel to the plane is equal to 2.4π multiplied by sin of 30 degrees. And the component perpendicular to the plane is equal to 2.4π multiplied by the cos of 30 degrees.

There are two forces acting parallel to the plane on body A, the tension force and the weight component, 2.4 π multiplied by sin 30. As the body is moving up the plane, the sum of the net forces is equal to π minus 2.4π multiplied by sin 30. This is equal to 2.4π. We know that sin of 30 degrees is equal to one-half. This means that the equation can be rewritten as π minus 1.2π is equal to 2.4π. We will call this equation one. For body B, we will resolve vertically. As the body begins to move downwards, we will consider this to be the positive direction. This means that the sum of the forces is equal to 1.6π minus π. This is equal to 1.6π, and we will call this equation two.

We will now clear some space so we can solve these two simultaneous equations to calculate the value of the acceleration π. Adding equation one and two eliminates π. This leaves us with 0.4π is equal to four π. We can then divide both sides of this equation by four. 0.1 multiplied by 9.8 is equal to 0.98. This means that the acceleration of the system is 0.98 meters per square second. We are told that 10 seconds later, the string breaks. We can use our equations of motion or SUVAT equations to calculate the velocity of the bodies at the point the string breaks. The bodies began at rest. They were accelerating at 0.98 meters per square second. And the string broke after 10 seconds. We will use the equation π£ is equal to π’ plus ππ‘.

Substituting in our values, we have π£ is equal to zero plus 0.98 multiplied by 10. This is equal to 9.8. Therefore, the velocity of the bodies at the point the string broke is 9.8 meters per second. At this point, body A will still move up the plane. When the string breaks though, the tension will be equal to zero. This means that the body will start to slow down. It will have a negative acceleration. To calculate this acceleration, we will use equation one. As π is equal to zero, negative 1.2π is equal to 2.4π. This means that π is equal to negative 0.5π. We divide both sides of the equation by 2.4. This gives us a value of π equal to negative 4.9 meters per square second.

We want to find the point where the body started to move in the opposite direction. This will happen immediately after the body comes to rest. Using our equations of motion once more, we know that π’, the initial velocity, is 9.8 meters per second. The final velocity is equal to zero meters per second. And the acceleration is equal to negative 4.9 meters per second squared. This means the body is decelerating at a rate of 4.9 meters per square second. Once again, we will use the equation π£ is equal to π’ plus ππ‘. Zero is equal to 9.8 plus negative 4.9 multiplied by π‘. Adding 4.9π‘ to both sides of the equation gives us 4.9π‘ is equal to 9.8. Finally, dividing through by 4.9 gives us π‘ is equal to two.

The time taken for the body to start moving in the opposite direction after the string broke is two seconds. This is a total of 12 seconds after the system was initially released.

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