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Lesson Explainer: Connected Particles Mathematics

In this explainer, we will learn how to solve problems with systems of multiple connected particles.

A freight train traveling on a railroad contains multiple freight cars that move as if they were one solid body on straight tracks. Connected particles act like a single particle with combined masses when moving in a straight path. In other words, we can mathematically model such motion either by considering the force acting on each connected particle separately or by assuming that the forces are acting on a single body with the combined masses of the connected particles. Both models lead to the same conclusion if the object does not change its direction of motion.

Of course, we need to first make idealistic assumptions about the connectors to make these two models truly equivalent. Freight cars are linked to each other by sturdy metal connectors. While these connectors are quite heavy, their mass is negligible in comparison to the mass of the freight cars they connect. Recall that a light object is an object whose mass is negligible or assumed to have zero mass. Hence, we often assume that the object connecting two particles is a light string or a light rod. We will generally use strings when the connector is pulling the other particle and rods when the connector is pushing the other particle.

In addition, we assume that the connectors are inextensible or incompressible. An inextensible string or incompressible rod means that the string or rod cannot be stretched or contracted regardless of the load that it bears. In other words, the length of an inextensible string or incompressible rod remains constant throughout the motion. Inextensible strings or incompressible rods are common modeling assumptions when the connector is used for pulling or pushing objects. These assumptions ensure that the distance between the connected particles remain constant so that the motion of individual connected components does not differ from that of the whole system. We summarize our assumptions in the following definition.

Definition: Light and Inextensible Connectors

Light and inextensible (or incompressible) connectors are strings (or rods) connecting two objects with zero mass that do not stretch (or contract) regardless of the load.

In our first example, we will find the acceleration of a particle in a connected system.

Example 1: Finding the Acceleration in a Connected Particles Problem

Two particles, A and B, are connected by a light and inextensible string and are lying on a flat rough surface. The masses of A and B are, respectively, 10 kg and 20 kg. A constant horizontal force of 300 N is acting on particle A away from particle B, causing both particles to move on a rough flat surface. During this motion, friction forces of magnitudes 50 N and 80 N are acting on particles A and B respectively. Find the magnitude of the acceleration of particle B, rounded to the nearest tenth of a metre per second squared.

Answer

We are given that the two particles are connected by a light and inextensible string. Here, “light” is a modeling assumption, meaning that we consider the mass of the string to be zero. Since it has zero mass, the string does not contribute additional mass to the connected-particle system. Also, “inextensible” means that it cannot be extended (stretched), so we can assume that the distance between A and B remains constant throughout the motion.

In order to find the acceleration of a particle, we recall Newton’s second law: forcemassacceleration=×.

Since we know the mass of both particles, we can find the acceleration by first finding the force acting on the particles. Let us begin by drawing a force diagram for the given context.

Since the motion is purely horizontal, we can ignore the vertical component of the force, which consists of the particles’ weight and the reaction of the horizontal surface on the weight. Also, since both particles are moving as one body, we can treat the forces as if acting on a single particle with the combined mass of two particles. To do this, we can first resolve the two friction forces so that the net friction force is given by 50+80=130.NNN

Using this assumption, we can draw a simpler force diagram.

Since we are only considering horizontal forces that are one dimensional in this context, we can associate the direction of motion with positive or negative signs. Let us say that the positive direction is toward the right side of the diagram. Then, the net horizontal force is given by 300130=170.NNN

This tells us that the resultant force acting on this particle is toward the right with a magnitude of 170 N. Since we have found the resultant force acting on this system, we can find the system’s acceleration by applying Newton’s second law. Since the mass of this single-particle system is the sum of the individual masses, we can obtain the mass to be 10+20=30.kgkgkg

Substituting the values for force and mass into Newton’s second law, 170=30×.Nkgacceleration

Dividing both sides of the equation by 30 kg, we obtain that the acceleration is 17030=5.̇6.Nkgms

This quantity is the acceleration of the connected-particle system and its positive sign indicates that it is to the right of the diagram. Since the two particles are connected by an inextensible string, where particle A is pulling behind particle B, the acceleration of particle B must be the same as that of the system. Hence, the acceleration of particle B is 5.7 m⋅s−2, rounded to the nearest tenth.

In the previous example, we saw that a connected-particle system is not so different from one-particle systems. As A starts moving, it pulls the string and this pulling force is transmitted to B via the string so that the string exerts the same pulling force on B that was exerted on it by A. We call this force tension, a word that comes from the Latin word tendere meaning “to pull taut”, which happens when the string is pulled from both ends. Tension is a new feature that arises in connected-particle systems.

Definition: Tension

Tension is a pulling force exerted on an object by a connector, such as a string or a rod. Due to Newton’s third law, which states that every action has an equal and opposite reaction, tension can also refer to a force exerted by an object on a connector.

Since tension is a type of force, it is a vector quantity, which has direction as well as magnitude. We can associate tension either with the object that the tension force is exerted on by a connector or with the connector on which the tension is exerted by one of the objects it is connecting. Depending on this choice, the direction of tension would be opposite but the magnitude would be the same, according to Newton’s third law. We can observe this in a diagram from the previous example.

Under idealistic assumptions, the magnitude of the tension is constant along the string, so the magnitude of the tension exerted on particle A by the string is the same as that exerted on particle B.

Sometimes, the term tension is also used to reference the magnitude of tension force experienced by the string. This is especially important for engineers designing connectors since they are interested in the maximum magnitude of tension force a connector can bear before breaking. In this lesson, however, we will use tension to refer to a vector force and magnitude of tension to refer to the scalar quantity.

It is easier to understand tension force in a connected-particle system if we think of the force diagram containing only one particle. Let us consider the force diagram containing only particle B from the previous example.

We can see that the string is exerting a horizontal force on particle B toward the right side of the diagram, which we call tension. With the positive direction to the right, the tension is thus a vector of components (𝑇,0), where 𝑇 is the magnitude of the tension. We can find the magnitude of the tension using particle B’s acceleration due to the resultant force, which we obtained previously. As discussed in the example, the vertical forces are balanced, so the resultant force is horizontal and its horizontal component is given by 𝑇80.N

Using particle B’s acceleration, 5.̇6 m⋅s−2, found in the example, Newton’s second law tells us that the resultant force on this particle must be equal to 20×5.̇6=113.̇3.kgmsN

Equating the two expressions for resultant force, we have 𝑇80=113.̇3,NN which gives us that the tension is 193.̇3 N. As we can see from this process, we only need to know one particle’s acceleration as well as the other force components acting on the particle in order to find the tension force exerted by the string.

In our next example, we will find the tension exerted by a string in a connected-particle system.

Example 2: Finding the Tension in the String in a Connected Particles Problem

Consider two particles, A and B, that are connected by a light inextensible string and are lying on a flat smooth surface. Particle A’s mass is 25 kg, while particle B’s mass is unknown. A constant horizontal force of 150 N is applied to particle B in the direction away from particle A, and both particle’s acceleration is horizontal with magnitude 5 m⋅s−2. Find the magnitude of the tension exerted by the string on particle A.

Answer

Let us begin be summarizing the modeling assumptions:

  • The string does not contribute additional mass to the system (as it is a light string).
  • The distance between both particles remains constant (because the string is inextensible, that is, it cannot be stretched), which means that both particles have the same motion.
  • The friction forces between the particles and the surface can be neglected because the particles slide on a smooth surface.

A force is applied to particle B away from particle A. As a result, particle B will move in the direction opposite to particle A, and the string will exert a pulling force on particle A, which is known as tension, and pull particle A with the same velocity. Let us draw a force diagram containing both particles.

Vertical forces represent the particle’s weight as well as the surface’s reaction to the weight. We can assume that these forces are balanced since the acceleration is horizontal. We can see in this diagram that the string is exerting a force on particle A toward the right direction on the diagram, which causes it to accelerate at 5 m⋅s−2 to the right direction.

We want to find the magnitude of tension exerted by the string. Recall that tension is a pulling force exerted by connectors such as strings. Tension is a vector quantity, and its direction depends on what object it is associated with. Here, the string connects particles A and B, where particle B is pulling particle A. Tension force is exerted by particle B, which is transmitted through the string to particle A. The direction of tension exerted by particle B is opposite to that exerted on particle A, but the magnitude of these two tensions are equal. Hence, we can solve this problem by considering the tension associated either with particle A or with particle B. We will present both methods below.

Method 1: First, let us compute the magnitude of tension acting on particle A. We begin by drawing a force diagram containing only particle A.

Since the vertical forces are balanced and friction forces are negligible, the resultant force acting on particle A is equal to the tension force. To find the magnitude of the resultant force, we recall Newton's second law: resultantforcemassacceleration=×.

We are given that particle A’s mass is 25 kg and that it is accelerating at 5 m⋅s−2. Substituting these values into the equation above, we obtain resultantforcekgmsN=25×5=125.

Hence, as the resultant force equals the tension, the magnitude of tension exerted by the string is 125 N.

Method 2: Let us compute the magnitude of tension exerted by particle B. We begin by drawing a force diagram containing only particle B.

Again, the vertical forces are balanced. Balancing horizontal forces gives us. 150𝑇, where 𝑇 is the magnitude of tension.

Now, to apply Newton’s second law, we need to know the mass of particle B. To find the mass, we can treat the connected-particle system as a one-particle system with the combined mass. Let us denote the mass of particle B by 𝑚 kg. Then, we know that a force of magnitude 150 N has resulted in an acceleration of magnitude 5 m⋅s−2. Then, Newton’s second law tells us (𝑚+25)×5=150.kgmsN

Rearranging the equation gives us 𝑚+25=1505=30𝑚=5.kg

Hence, the mass of particle B is 5 kg.

Returning to the problem at hand, the horizontal component of the resolvent force has led to an acceleration of magnitude 5 m⋅s−2 for particle B with mass 5 kg. Applying Newton’s second law, we obtain (150𝑇)=5×5.Nkgms

Rearranging the equation gives us 150𝑇=25𝑇=15025=125,N which agrees with the previous conclusion.

Hence, the magnitude of tension exerted by the string is 125 N.

In the previous example, we calculated the magnitude of tension force exerted by a string. We can easily see the importance of such a concept in real life when we think of a freight train, which we discussed at the start of this explainer. A pair of freight cars are connected by a metal connector that exerts force required for the next car to move at the same velocity. This connector plays a very important role in the train and bears a tremendous amount of force. If the connector were to fail, the following car would be disconnected from the front of the train, which can lead to very dangerous accidents. Hence, it is important to know first how much tension force the connector is expected to encounter so that the connector can be designed accordingly.

In the next example, we will consider a real-world connected particle problem.

Example 3: Solving a Real-World Connected Particle Problem

An engineer is designing a connector for a pickup truck to be used when the truck needs to pull a trailer behind. To ensure the safety of the connector, the engineer needs to calculate the maximum tension that the connector must be able to handle without breaking. The truck’s maximum mass with a driver and passengers is 1‎ ‎500 kg, and the maximum mass of the trailer with a full load is assumed to be 5‎ ‎000 kg. At the maximum capacity, the truck and trailer will experience horizontal friction of magnitudes 300 N and 1‎ ‎000 N respectively. Assuming that a connector is a light inextensible string connecting the truck and the trailer, find the magnitude of the tension in the connector under maximum capacity when the truck is on a horizontal road accelerating at 0.5 m⋅s−2.

Answer

Recall that tension is a pulling force exerted by a connector. In this example, we are given that a connector, connecting the truck and the trailer, is modeled by a light inextensible string. “Light” means that we can ignore the mass of the connector, and “inextensible” means that the connector will not stretch or break regardless of the load. Let us begin by drawing a force diagram.

Vertical forces represent the weight of the truck and the trailer as well as the surface’s reaction to the weight. We can assume that these forces are balanced since the acceleration is horizontal. The driving force to the right sets the truck into motion, which pulls the string, and this pulling force is then transmitted to the trailer via the string, causing the truck and trailer to accelerate at 0.5 m⋅s−2 to the right direction.

We want to find the tension in the connector when the truck and trailer are at full capacity. Recall that the tension in the connector (that is, the tension exerted on the connector by either the truck or the trailer) has the same magnitude as but opposite direction to the tension exerted by the connector on either the truck or the trailer. Hence, let us consider a force diagram showing the trailer only so that the tension will be considered.

Since the vertical forces are balanced, the resultant force acting on the trailer is given by 𝑇1000,N where 𝑇 is the magnitude of tension and the positive direction is to the right side of the diagram. We are given that the truck’s, hence the trailer’s, acceleration is equal to 0.5 m⋅s−2 and also that the mass of the trailer under maximum load is 5‎ ‎000 kg. Newton’s second law tells us that the magnitude of the resultant force on this particle must be equal to 5000×0.5=2500.kgmsN

Equating the two expressions for resultant force, we have 𝑇1000=2500.NN

Adding 1‎ ‎000 N to both sides of the equation, we obtain that the magnitude of tension exerted by the connector under the maximum capacity, when the truck is accelerating at 0.5 m⋅s−2, is 3‎ ‎500 N.

So far, we have considered systems where particles were connected by a light inextensible string. Let us turn our focus to a system where particles are in contact with each other. For instance, we can think of two heavy boxes stacked vertically that are lifted by a forklift truck. In this case, there is no connector between the two objects in motion. The forklift truck exerts force on the bottom box, which leads to an upward acceleration. The bottom box exerts force on the top box so that the top box experiences the same acceleration.

During this acceleration, the bottom box is also experiencing additional force from the box on top. To better understand this force, we recall Newton’s third law: Every force has an equal and opposite reaction force.

Let us visualize this principle using a force diagram.

In the diagram above, each force vector is labeled by the object exerting the force. As mentioned previously, the forklift exerts an upward force on the bottom box, which leads to the acceleration of the bottom box. The corresponding downward force is the reaction to this upward force. The reaction, in this case, is the downward force exerted by the bottom box on the forklift truck. The direction of the reaction is opposite to that of the original force, and the magnitude of the reaction is equal to that of the original force.

We can also observe the reaction of the top box on the upward force exerted by the the bottom box. The reaction of the top box is the downward force equal in magnitude to the upward force that the bottom box exerts on the top box.

In the next example, we will calculate the reaction force where two particles are in contact with each other.

Example 4: Finding the Force Exerted by a Mass on Top of Another Mass

A 500 g mass is lying on top of a 2 kg mass while the bottom mass is accelerating upward with magnitude 2 m⋅s−2. If the positive direction is upward, find the force that the 500 g mass is exerting on the 2 kg mass to the nearest tenth of a newton.

Answer

Since the mass on the bottom is accelerating upward while the object on top (that is, the 500 g mass) is lying on top of it, the object on top must have the same acceleration. To cause this acceleration, the bottom object (that is, the 2 kg mass) must be exerting force on the top object. However, the problem asks us to find the force exerted by the top object (that is, the 500 g mass). In order to answer this question, we need to recall Newton’s third law: every force has an equal and opposite reaction.

We can visualize this principle using a diagram.

The upward force indicated in the diagram is the force exerted by the bottom object on the top object, causing it to accelerate upward at 2 m⋅s−2. The downward force is equal in magnitude to this upward force and opposite in direction. This is the reaction of the top object implied by Newton’s third law. Hence, this diagram reveals that the force exerted by the 500 g mass will be downward. Since the magnitude of this force is equal to that of the upward force, it suffices to identify the upward force acting on the top object using the given acceleration. Let us draw a force diagram containing only the 500 g mass.

The weight of this object can be calculated by multiplying its mass by the gravitational acceleration, which is equal to 9.8 m⋅s−2. Here, we need to convert the mass to kilograms, that is, 0.5 kg:weightkgmsN=0.5×9.8=4.9.

Let us assume that upward is the positive direction in this diagram. Denoting the unknown force by 𝐹 N, the resultant force on this object is equal to (𝐹4.9) N.

We recall Newton’s second law: forcemassacceleration=×.

Since we are given that this object is accelerating upward at 2 m⋅s−2, the resultant force should also equal to 0.5×2=1.kgmsN

Equating both expressions for the resultant force, we obtain 𝐹4.9=1, which leads to 𝐹=5.9N. This gives the magnitude of the upward force, which is the force exerted by the 2 kg mass on the 500 g mass. Using Newton’s third law, the force that the 500 g mass is exerting on the 2 kg mass should be equal in magnitude but opposite in direction.

Hence, the 500 g mass is exerting a 5.9 N downward force on the 2 kg mass.

In our final example, we will solve another reaction problem where a forklift truck is lifting vertically stacked boxes.

Example 5: Finding the Force Exerted by a Mass on Top of Another Mass

Two boxes are stacked up vertically on a forklift truck. The mass of the box on top is 200 kg, while the mass of the box at the bottom is 600 kg. The forklift truck applies a constant upward force of 10‎ ‎000 N, lifting both boxes. Find the magnitude of the force the top box exerts on the bottom box.

Answer

As the forklift exerts an upward force on the bottom box, the bottom box exerts an upward force on the top box. The whole system will be set into motion if the magnitude of the force exerted by the forklift is larger than the weight of the two boxes. Since the two boxes move in the same straight line, they can be treated as a single body, which means that they have the same acceleration, given by Newton’s second law. Once we have the acceleration, we will be able to find all forces exerted on the top box. This will allow us to find the force exerted by the top box on the bottom box using Netwon’s third law, which states that if body A exerts force on body B, then the force exerted by B on A is equal in magnitude but opposite in direction.

Let us start by finding the acceleration of the whole system.

We are given that the forklift truck is exerting an upward force of magnitude 10‎ ‎000 N. Recall that Newton’s second law tells us that forcemassacceleration=×.

Let us draw a force diagram of the whole system.

The mass of the whole system, made of the two boxes, is given by 200+600=800.kgkgkg

The weight of the whole system can be calculated by multiplying its mass by the gravitational acceleration, which is equal to 9.8 m⋅s−2: weightkgmsN=800×9.8=7840.

Assuming the positive direction to be toward the top side of the diagram, the resultant force on the whole system is equal to 100007840=2160.NNN

We can substitute this resultant force, along with the particle’s mass of 800 kg, to find the acceleration of this particle: 2160=800×.Nkgacceleration

Dividing both sides of the equation by 800 kg, we find accelerationNkgms=2160800=2.7.

This is the acceleration of the whole system, which is also the acceleration of each box. In particular, the top box is accelerating upward at 2.7 m⋅s−2. Now, we can use this information to find the magnitude of the force the bottom box is exerting on the top box. Let us draw a force diagram for the top box.

Weight of the top box can be calculated as previously but using only the mass of the top box, which is 200 kg: weightkgmsN=200×9.8=1960.

Again, we assume that upward is the positive direction in this diagram. Denoting the unknown force by 𝐹 N, the resultant force on this object is equal to (𝐹1960) N.

On the other hand, since we know the acceleration of this object, we can apply Newton’s second law to find another expression for the resultant force. Using the previously obtained value for acceleration, 2.7 m⋅s−2, the resultant force should also equal 200×2.7=540.kgmsN

Equating both expressions for the resultant force, we obtain 𝐹1960=540, which leads to 𝐹=2500N.

Hence, the magnitude of the force the top box is exerting on the bottom box is 2‎ ‎500 N.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • When connected particles are moving in the same straight line, we can treat them as a single particle of mass equal to the combined masses of individual particles.
  • Tension is a pulling force exerted on an object by a connector, such as a string or a rod. To calculate a tension force, it is beneficial to consider the force diagram containing only the object on which the tension force is exerted.
  • Newton’s third law states that every force has an equal and opposite reaction force.
  • We can find the reaction force by first identifying the original force corresponding to the reaction and then reversing the direction.

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