A body of mass 6.8 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 to the top of the plane to another body of mass 5.1 kilograms hanging freely below the pulley. When the system was released from rest, the two bodies were on the same horizontal level. Determine the vertical distance between the two bodies two seconds after they started moving. Take 𝑔 equal to 9.8 meters per second squared.
We will begin by sketching a diagram where we model the two bodies as particles. We are told that a body of mass 6.8 kilograms rests on a smooth inclined plane. This means that there is a force acting vertically downwards equal to its weight. And this is equal to its mass of 6.8 kilograms multiplied by gravity. We are told to take this value of 𝑔 equal to 9.8 meters per second squared.
As the plane is smooth, there will be no frictional force. And we are told that the angle of inclination is 30 degrees. The body is connected via a light inextensible string to another body of mass 5.1 kilograms. This hangs freely below a pulley and exerts a downward force equal to 5.1 kilograms multiplied by 𝑔. Since the pulley is smooth, the tension in the string will be constant throughout. And since the string is inextensible, when the system is released, the acceleration will also be constant. When the system was released, we are told that the two bodies were on the same horizontal level. Since we are modeling the bodies as particles, we don’t need to worry about the center of gravity of the objects.
We have been asked to calculate the vertical distance between the two bodies two seconds after they started moving. We can do this using the equations of motion or SUVAT equations. But we firstly need to calculate the acceleration of the system 𝑎.
Recalling Newton’s second law, we know that the sum of the forces acting on a body are equal to its mass multiplied by acceleration. As the freely hanging body will accelerate vertically downwards, we will let this direction be positive. And the sum of the forces is therefore equal to 5.1𝑔 minus 𝑇. This is equal to 5.1 multiplied by 𝑎.
The weight force acting on the body on the plane can be split into components acting parallel and perpendicular to the plane. Using our knowledge of right angle trigonometry, these are equal to 6.8𝑔 multiplied by sin of 30 degrees and 6.8𝑔 multiplied by the cos of 30 degrees. It is the component of the weight force acting parallel to the plane that we are interested in. Since this body accelerates up the plane, we will let this be the positive direction. And the sum of our forces is equal to 𝑇 minus 6.8𝑔 multiplied by sin of 30 degrees. We know that sin of 30 degrees is equal to one-half. So our expression simplifies to 𝑇 minus 3.4𝑔, and this is equal to 6.8𝑎.
We now have a pair of simultaneous equations that we can solve by elimination. Adding equation one and two, we have 1.7𝑔 is equal to 11.9𝑎. Dividing through by 11.9, we have 𝑎 is equal to 1.4. The acceleration of the system is therefore equal to 1.4 meters per second squared.
Whilst we could substitute this value of 𝑎 into equation one or equation two to calculate the tension force 𝑇, this is not required in this question. Instead, we will now calculate the displacement of the bodies in the first two seconds of motion. We know that the initial velocity of the bodies is zero meters per second. They accelerate at 1.4 meters per second squared. And we need to calculate the displacement 𝑠 after two seconds. We will use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. And substituting in our values, we have 𝑠 is equal to zero multiplied by two plus a half multiplied by 1.4 multiplied by two squared. This is equal to 2.8. The displacement of the bodies after two seconds is therefore equal to 2.8 meters.
In order to calculate the distance between the two bodies, we might be tempted to double this answer. The freely hanging body has fallen a distance of 2.8 meters. However, the body on the slope has moved a distance of 2.8 meters up the plane. And we are interested in the vertical distance it has traveled. By sketching a right triangle and using the sine ratio, we see that the sin of 30 degrees is equal to 𝑦 over 2.8, where 𝑦 is the vertical distance traveled by this body. Once again, we recall that sin of 30 degrees is one-half. And multiplying through by 2.8, we have 𝑦 is equal to 1.4 meters.
We are now in a position to calculate the vertical distance between the two bodies. This is equal to 2.8 meters plus 1.4 meters. And we can therefore conclude that the vertical distance between the two bodies two seconds after they started moving is 4.2 meters.