Video: Applications of Newton’s Second Law: Inclined Pulley

In this video, we will learn how to solve problems on the motion of two bodies connected by a string passing over a smooth pulley with one of them on an inclined plane.

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Video Transcript

In this video, we will learn how to solve problems on the motion of two bodies connected by a string passing over a smooth pulley, with one of them on an inclined plane. We will study problems where the plane is both smooth and rough. To begin this video, we will recall Newton’s second law of motion.

Newton’s second law states that the net force on an object is equal to its mass multiplied by its acceleration. In this video, we will be dealing with scalar quantities and use the equation 𝐹 equals 𝑚𝑎, where 𝐹 is the sum of the net forces, 𝑚 is the mass, and 𝑎 the acceleration. We will measure acceleration in meters per square second. The mass will be measured in kilograms. Multiplying these gives us kilogram meters per square second. One kilogram meter per square second is equal to one newton. And this is the unit we’ll use to measure our forces.

We will now consider what the problem will look like when the plane is smooth. We will have two bodies 𝐴 and 𝐵 connected by a light inextensible string passing over a smooth pulley. This means that the tension in the string will be equal throughout. Both bodies will have a force going vertically downwards equal to its weight. This will be equal to the mass of the body multiplied by gravity. There will be a reaction force perpendicular to the plane. However, as the plane is smooth, there will be no frictional force on it.

As the string is inextensible, when the system is released, it will accelerate uniformly. We will then use Newton’s second law to resolve for body 𝐴 and body 𝐵. In the case of body 𝐵, we will resolve vertically. For body 𝐴, we will resolve parallel to the plane. In order to do this, we will need to know the angle of inclination, in this case 𝛼. Using our knowledge of right angle trigonometry, we can identify the component of the weight of body 𝐴 parallel and perpendicular to the plane. The force acting parallel to the plane is equal to 𝑀𝑔 multiplied by sin 𝛼, and the force perpendicular to the plane is equal to 𝑀𝑔 multiplied by cos 𝛼.

If body 𝐵 is accelerating downwards and we consider this to be the positive direction, the sum of the forces are equal to 𝑀𝑔 minus 𝑇. This is equal to the mass of body 𝐵 multiplied by the acceleration. Body 𝐴 is moving up the plane. If this is the positive direction, the sum of the forces acting parallel to the plane are equal to 𝑇 minus 𝑚𝑔 sin 𝛼. This is equal to the mass of body 𝐴 multiplied by the acceleration.

This will give us two simultaneous equations that we can use to calculate any unknowns. We will now look at a couple of examples.

A body of mass five kilograms rests on a smooth plane inclined at an angle of 35 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 19 kilograms hanging freely vertically below the pulley. Given that the acceleration due to gravity 𝑔 is equal to 9.8 meters per square second, determine the acceleration of the system.

We will begin by sketching the system. We are told that the smooth plane is inclined at an angle of 35 degrees. The masses of the two bodies are five kilograms and 19 kilograms. This means that they will have a vertical downward force equal to five 𝑔 and 19𝑔, respectively, where the gravity 𝑔 is equal to 9.8 meters per square second.

The pulley is smooth, and the string is light and inextensible. This means that the tension in the string will be equal throughout. When the system is released, the magnitude of acceleration will also be constant. As the plane is smooth, there will be no frictional force. In order to solve this problem, we will use Newton’s second law, which states that the sum of the net forces is equal to the mass multiplied by the acceleration. For the body on the plane, we will resolve parallel to the plane. And for the body hanging freely, we will resolve vertically.

The five 𝑔 force is neither parallel nor perpendicular to the plane. Therefore, we need to calculate these two components. Using our knowledge of right angle trigonometry, we see that the perpendicular component is equal to five 𝑔 multiplied by cos of 35 degrees and the component parallel to the plane is equal to five 𝑔 multiplied by sin of 35 degrees.

Body 𝐴 is moving up the plane. If we assume this is the positive direction, the sum of its forces is equal to 𝑇 minus five 𝑔 multiplied by sin 35. This is equal to five 𝑎 as the mass of the object is five kilograms. As object 𝐵 is accelerating downwards, the sum of its forces is equal to 19𝑔 minus 𝑇. This is equal to 19𝑎. We now have two simultaneous equations that we can solve to determine the acceleration of the system. By adding equation one and equation two, we eliminate the tension 𝑇.

The left-hand side becomes 19𝑔 minus five 𝑔 multiplied by sin of 35 degrees. And the right-hand side becomes 24𝑎. We can then divide both sides of this equation by 24. Typing this into the calculator gives us a value of 𝑎 equal to 6.5872 and so on. Rounding this to two decimal places gives us an answer of 6.59 meters per square second. We could substitute this value back into equation one or equation two to calculate the tension 𝑇. However, this is not required in this question.

We will now consider what happens when the surface is rough. Recall in the diagram we saw earlier for a smooth plane we used Newton’s second law to create two equations. We resolved vertically for the freely hanging object and resolved parallel to the plane for the object on the inclined plane. Let’s now consider what happens when the plane is rough.

When body 𝐴 is accelerating up the plane, there will be a frictional force acting downwards parallel to the plane. This means that we have a tension force acting in the positive direction and two forces acting in the negative direction. The sum of the net forces parallel to the plane is equal to 𝑇 minus 𝑚𝑔 multiplied by sin 𝛼 minus the frictional force 𝐹 𝑟. This is equal to the mass multiplied by the acceleration.

We know that when dealing with a frictional force, it is equal to 𝜇, the coefficient of friction, multiplied by the normal reaction force 𝑅, when 𝜇 is a constant that lies between zero and one inclusive. If we resolve perpendicular to the plane, we see that the sum of the net forces is equal to 𝑅 minus 𝑚𝑔 multiplied by cos 𝛼. However, the body is not moving in this direction. Therefore, the acceleration is equal to zero. 𝑅 minus 𝑚𝑔 multiplied by cos 𝛼 is equal to zero.

This can be rewritten as 𝑅 is equal to 𝑚𝑔 multiplied by cos 𝛼. When solving any problem of this type on a rough plane, we can use our three equations to calculate any unknowns, together with the formula that links the frictional force and the normal reaction force. We will now look at an example of this type.

A body 𝐴 of mass 240 grams rests on a rough plane inclined to the horizontal at an angle whose sine is three-fifths. It is connected by a light inextensible string passing over a smooth pulley fixed to the top of the plane to another body 𝐵 of mass 300 grams. If the system was released from rest and body 𝐵 descended 196 centimeters in three seconds, find the coefficient of friction between the body and the plane. Take 𝑔 equal to 9.8 meters per square second.

We begin by sketching the system. We are told that the sin of angle 𝛼 is equal to three-fifths. Using our knowledge of right angle trigonometry and the Pythagorean triple three, four, five, we know that cosine of 𝛼, or cos of 𝛼, is equal to four-fifths. The mass of the two bodies is given in grams. We know that there are 1000 grams in one kilogram. This means that 240 grams is equal to 0.24 kilograms. We divide our value in grams by 1000.

Body 𝐴 will, therefore, have a force acting vertically downwards equal to 0.24𝑔, where 𝑔 is equal to 9.8 meters per square second. Body 𝐵 has a mass of 300 grams, and this is equal to 0.3 kilograms. Therefore, this body has a downward force of 0.3 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 when released, the system will travel with uniform acceleration. Body 𝐴 has a normal reaction force perpendicular to the plane. As the plane itself is rough, there will be a frictional force acting down 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, to resolve parallel and perpendicular to the plane for body 𝐴 and vertically for body 𝐵. The weight of body 𝐴 is acting vertically downwards. Therefore, we need to find the components of this that are parallel and perpendicular to the plane. Once again, using our knowledge of right angle trigonometry gives us a force of 0.24𝑔 multiplied by cos 𝛼 perpendicular to the plane and 0.24𝑔 multiplied by sin 𝛼 parallel to the plane.

There are three forces acting on 𝐴 parallel to the plane: the tension force, the frictional force, and this weight component. This gives us the equation 𝑇 minus 0.24𝑔 multiplied by sin 𝛼 minus the frictional force 𝐹 𝑟 is equal to 0.24𝑎. Perpendicular to the plane, the sum of the net forces is equal to 𝑅 minus 0.24𝑔 multiplied by cos 𝛼. The body is not moving in this direction. Therefore, this is equal to zero.

Rearranging the equation, we see that the normal reaction force 𝑅 is equal to 0.24𝑔 multiplied by cos 𝛼. Finally, resolving vertically for body 𝐵, where downwards is the positive direction, gives us 0.3𝑔 minus 𝑇 is equal to 0.3𝑎. We can substitute in our values for sin 𝛼 and cos 𝛼. This means that the normal reaction force is equal to 1176 over 625. 0.24𝑔 multiplied by sin 𝛼 is equal to 882 over 625.

Our next step is to calculate the acceleration of the system, given the fact that body 𝐵 descended 196 centimeters in three seconds. In order to do this, we will use the equations of motion, also known as the SUVAT equations. The displacement of the body was 196 centimeters. This is equal to 1.96 meters, as there are 100 centimeters in a meter. The body was released from rest, so the initial velocity is zero meters per second. We are trying to calculate the acceleration, and we are told the time is three seconds.

This means that we can use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us 1.96 is equal to zero multiplied by three plus a half multiplied by 𝑎 multiplied by three squared. The right-hand side simplifies to 4.5𝑎. Dividing both sides by 4.5 gives us 𝑎 is equal to 98 over 225. The acceleration of the system is 98 over 225 meters per square second. We can now substitute this value into our equations.

We now have two unknowns left, the tension 𝑇 and the frictional force 𝐹 𝑟. We know that the normal reaction force is 1176 over 625 newtons. By adding 𝑇 and subtracting 49 over 375 to both sides of the bottom equation, we can calculate the tension force 𝑇. This gives us a value of 𝑇 equal to 2107 over 750 newtons. We can substitute this into the top equation. Rearranging this equation gives us a frictional force 𝐹 𝑟 equal to 1617 over 1250 newtons.

We now need to calculate the coefficient of friction. And we know that the frictional force is equal to this coefficient of friction multiplied by the normal reaction force. This means that the coefficient of friction 𝜇 is equal to 𝐹 𝑟 divided by 𝑅. Typing this into the calculator gives us a coefficient of friction 𝜇 equal to eleven sixteenths.

We will now summarize the key points from this video. To solve problems involving pulleys on an inclined plane, we used Newton’s second law, 𝐹 equals 𝑚𝑎. We resolve forces vertically as well as parallel and perpendicular to the plane. If the plane is rough, we have a frictional force 𝐹 𝑟 acting against the motion, where this frictional force is equal to the coefficient of friction 𝜇 multiplied by the normal reaction force 𝑅. We can also use the equations of motion, or SUVAT equations, to calculate unknowns and help us solve problems of this type.

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