# Lesson Explainer: Applications of Newton’s Second Law: Two Masses Hanging from a Pulley Mathematics

In this explainer, we will learn how to solve problems on the motion of a system of two bodies suspended vertically by a string passing over a smooth pulley.

Consider two uniform rectangular cuboid bodies connected by a string that passes over a pulley that is fixed below a surface, as shown in the following figure. The lengths of the bodies into the plane of the page are equal. The forms of the bodies are specified so that the larger body shown is necessarily also the body with greater mass.

If the string is light and inextensible, and the force required to make the pulley rotate is negligible, the acceleration of either of the bodies depends only on the weights of the bodies and the tension in the string.

The fact that the string connecting the bodies is inextensible implies that the distance moved by one of the bodies in a time interval must equal the distance moved by the other body. From this, we see that the changes in speed of both bodies in a time interval must be equal. The accelerations of either body must then have equal magnitude. The value of the magnitude of the acceleration of the bodies can be determined by considering the magnitudes and directions of the weights of the bodies and the tension in the string.

The forces acting on the bodies are shown in the following figure.

The body with mass is the larger body. As the bodies are uniform, it must be the case that

As it is the case that and the magnitude of acting vertically upward on each body is the same, it must be the case that

The net forces on the bodies are shown in the following figure.

The forces acting will result in the body of greater mass accelerating downward and the body of lesser mass accelerating upward.

We can see that the net force on the larger body is greater than the net force on the smaller body. The difference between the net forces depends on the value of the difference between the masses of the bodies, given by

The net force on the larger body does not produce a greater acceleration, as the mass of the larger body is greater by the same value of the difference of the masses of the bodies that produces the unequal forces on them.

The force acting on and the acceleration of a body are related by Newton’s second law of motion. Let us define this.

### Definition: Newton’s Second Law of Motion

When a net force, , acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body, according to the formula where is the mass of the body and is the acceleration of the body.

We can now use Newton’s second law of motion to analyze the motion of the bodies suspended by the string.

The body of mass is the body with greater mass, so the tension in the string acting upward must be less than the weight of the body of mass acting downward. The magnitude of the net force on the body is given by

For the body of lesser mass, , the tension in the string acting upward must be greater than the weight of the body of mass acting downward. The force on the body is given by

For each body, the net force on the body is related to the acceleration of the body by the formula so we can see that and

These expressions can be rearranged as and

Adding these expressions allows us to eliminate :

We can then rearrange to make the subject: where is acceleration due to gravity.

In the case that equals , we see that must equal . In this case, the equation has a numerator equal to zero and the acceleration of either body is zero.

Let us look at an example of two bodies suspended vertically from a pulley.

### Example 1: Calculating the Acceleration of a System of Two Bodies Suspended over a Smooth Pulley

Two bodies of masses 12 kg and 18 kg are attached to the ends of a light inextensible string which passes over a smooth pulley. Determine the acceleration of the system. Take .

### Answer

The acceleration of either body is given by the formula where is the mass of the body with the greater mass and is the mass of the body with the lesser mass.

Substituting known values into the formula gives us

Let us now look at an example of two bodies suspended vertically from a pulley where the mass of one of the bodies is unknown.

### Example 2: Calculating an Unknown Mass in a System of Two Bodies Suspended over a Smooth Pulley

Two masses and 88 g are attached to the ends of a light string passing over a smooth pulley. Determine the value of , given that, when the system was released, the other mass descended 11.76 m in 2 seconds. Take the acceleration due to gravity .

### Answer

The acceleration of either body is given by the formula where is the mass of the body with the greater mass and is the mass of the body with the lesser mass. As the 88-gram mass descends, it has the greater mass.

Substituting known values into the formula gives us

The value of can be determined using the formula

The value of is zero, as the body accelerates from rest. The formula must be rearranged to make the subject, so

Substituting the values of and , we obtain

Substituting the value of determined, we have

Multiplying both sides of the equation by the denominator of the fractional term, we have

Expanding brackets gives us

The unknown mass is 22 grams.

It has been shown that the acceleration of a system of two bodies with masses and , suspended by a pulley, is given by

It has also been shown that, for each of the bodies, the net force on the body is related to the acceleration of the body by the formula

Hence, it is the case that and

Either of these expressions can be rearranged to make the subject, as follows: and

Let us look at an example where the tension in the string connecting two bodies suspended by a pulley is determined.

### Example 3: Solving a Multistep Problem Involving a System of Two Bodies Suspended over a Smooth Pulley

Two bodies of masses 270 and grams are connected by two ends of a string passing over a smooth pulley. The body of mass was projected downward at 105 cm/s and, 3 seconds later, it returned to its initial position. Find the value of and the tension in the string. Take .

### Answer

The body of mass is projected downward, but the force acting on it must be upward as it returns to its initial position.

The body of mass descends and then returns to its starting point in a time interval of 3 seconds. The acceleration of the body is given by the formula

As the acceleration of the body is uniform, the body is instantaneously at rest at the midpoint of the 3-second time interval, so at the midpoint of the 3-second time interval. Taking the upward direction as positive, the acceleration is given by

The tension in the string provides the acceleration that decreases the downward velocity of the body of mass ; hence, and the acceleration of either body is given by

The value of used is in m/s2, so it is convenient to convert the value of to a value in m/s2, giving us

The tension in the string is given by where is the mass of the body with the greater mass, in this case 270 grams. To obtain a result in newtons, the mass is converted into a mass in kilograms of 0.27 kg. The tension is, therefore, given by

Let us now look at an example where two bodies are suspended vertically from a pulley where the connection between the bodies breaks while the bodies are in motion.

### Example 4: Solving a Multistep Problem Involving a System of Two Bodies Suspended over a Smooth Pulley and a Broken String

Two bodies of masses 374 g and 102 g were connected to each other by a light inextensible string passing over a smooth pulley. The two bodies started at rest on the same horizontal level. Then one second after the system was released, the string broke. Determine the vertical distance between the two bodies one second after the string broke. Take the acceleration due to gravity .

### Answer

The vertical distance between the bodies one second after the string breaks depends on the motions of the bodies in the time between their being at rest and the instant one second after the spring breaks.

The following figure represents the positions of the bodies at one-second intervals, starting at the instant when they are released. The second diagram shows when the string breaks, and the third diagram shows one second after the string breaks (the diagram is not to scale).

The velocities of the bodies at each instant are shown in red. The displacements of the bodies from their positions one second earlier are shown in blue.

The arrow representing has been drawn with a head at each end. This has been done because the body of mass 102 grams is accelerating upward at the instant that the string breaks and downward after the string breaks. It is not yet determined whether one second after the string breaks the body will have an upward or a downward displacement.

The final vertical distance between the bodies is represented by the following figure, in which orange displacements are downward and green displacements are upward.

The arrow representing has been drawn with a head at each end and as both green and yellow, as its direction has not been determined.

Before the string breaks, either body has an acceleration with a magnitude given by

Let us first consider the motion of the body of mass 374 grams.

For one second, the body accelerates vertically at 5.6 m/s2. At this instant, the vertical displacement of the body is given by

The downward direction is taken as positive.

The vertical speed of the body at this instant is given by

When the string breaks, the body accelerates vertically downward at 9.8 m/s2. After one second of acceleration, the vertical displacement of the body is given by

Now let us consider the motion of the body of mass 102 grams.

For one second, the body accelerates vertically at m/s2. At this instant, the vertical displacement of the body is given by

The vertical speed of the body at this instant is given by

When the string breaks, the body accelerates vertically at 9.8 m/s2. After one second of acceleration, the vertical displacement of the body is given by

The negative value of shows that the 102-gram mass reaches a point vertically above the position it had when the string broke.

We can now determine the displacement of the bodies from each other. This is given by the displacement of the body of greater mass, , minus the displacement of the body of lesser mass, :

The vertical distance between the bodies is 16.8 metres, which is 1‎ ‎680 centimetres.

Now let us look at an example of two bodies suspended vertically from a pulley where the masses of the bodies are unknown but the difference between the masses of the bodies is known.

### Example 5: Calculating the Force Exerted on a Pulley Given a System of Two Bodies Suspended over a Smooth Pulley

Two bodies of masses g and g are connected to each other by a light string which passes over a fixed smooth pulley. The system was released from rest when the two bodies were at the same horizontal level. One second later, the vertical distance between them was 128 cm. Find the magnitude of the force exerted on the pulley while the bodies were in motion. Take the acceleration due to gravity .

### Answer

A tension exists in the string. Each part of the string that supports a body exerts on the pulley a vertically downward force equal to the tension in the string on the pulley. The tension in the string must, therefore, be determined to determine the force on the pulley. We assume that the weight of the pulley is negligible so that only the tension forces act on the pulley.

The masses of the bodies are unknown, so the tension in the string cannot be directly determined. The vertical distance between the pulleys after one second of acceleration allows us however to determine the acceleration of either body. The relation between the acceleration of either body is related to the difference in the masses of the bodies and to the tension in the string, allowing the tension to be determined.

The following figure shows the forces on the bodies and their accelerations at the instant that the bodies are released and also the vertical distance between the bodies one second after they are released.

The accelerations of the bodies have equal magnitude, so the distances moved by each body in one second are equal, and each must be equal to half of the vertical distance between the bodies. The magnitude of the displacement of either body in one second is given by

Hence,

The acceleration of either body is given by

The vertical distance is given as a value in centimetres and the time in which the bodies move is given as a value in seconds; hence, the acceleration of the system is 128 cm/s2. To be consistent with this value, the acceleration due to gravity will also be given in units of cm/s2 and is, therefore, 980 cm/s2. We see then that

This expression can be rearranged to make the subject as follows:

If we take the body of mass , we know that the tension in the string connected to it is given by

Hence,

Substituting the value of obtained, we have that

The tension is a force, which in this case has been calculated by using values of mass with the unit of grams and values of acceleration with the unit of cm/s2, so the unit of the force calculated is g⋅cm/s2, which is the definition of the unit dyne.

Each part of the string exerts a force of 206‎ ‎503.5 dynes on the pulley, so the force on the pulley from the two strings is given by

Let us summarize what has been learned in these examples.

### Key Points

• For two bodies suspended vertically by a light, inextensible string running over a smooth pulley that requires negligible force to be turned, the acceleration of either body is given by where and are the masses of the bodies and is acceleration due to gravity.
• The tension in the vertical string connecting the bodies of masses and is given by where For the case that the acceleration of either body is zero.

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