7.6: Dependent Motion Systems

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Dependent motion analysis is used when two or more particles have motions that are in some way connected to one another. The way in which the motion of these particles is connected is known as the constraint. A simple example of a constrained system is shown in the figure below. Imagine that someone's pickup truck gets stuck in the sand, and a friend uses a rope to help pull it out. This friend ties one end of the rope to the rear bumper of her car, loops the rope around a bar on the front of the pickup truck, then ties the other end to a stationary tree. In this case the two vehicles will not have the same velocity or acceleration, but their motions are related because they are tied together by the rope. In this case, the rope is acting as the constraint, allowing us to know the velocity or acceleration of one vehicle based on the velocity or acceleration of the other vehicle.

A pickup truck is stuck in sand to the left side of a tree, and a car is on the right side of the tree. A rope has one end tied to the car's rear bumper, is looped through an attachment point on the front of the truck, and has its other end tied to the trunk of the tree. The rope is taut and horizontal throughout. The distance between the tree and the truck's front is labeled L_A, and the distance between the tree and the car's rear is labeled L_B. — Figure \(\PageIndex{1}\): This represents a constrained system. The motion of the car and the pickup truck will be related to one another via rope that is connecting them.

The first thing we will need to do when analyzing these systems is to come up with what is known as the constraint equation. A constraint equation will be some geometric relationship that will remain true over the course of the motion. In the above example, imagine the rope is 50 feet long. Using the tree as a stationary point, we can also say that the length of the rope is the distance from the green car to the tree, plus two times the distance from the pickup truck to the tree (since it must go out to the pickup truck and then back). If we put this into an equation (the constraint equation) we would have the following.

\[ \text{Constraint Equation: Positions} \quad \,\, \, L = 50 \, ft = 2 L_A + L_B \]

Once we have a constraint equation that works for positions, we can take the derivative of this equation to find another constraint equation that relates velocities. In this case, the length of the rope is constant, and therefore the derivative of the length will be zero. If we take the derivative of the constraint equation again, we wind up with a third constraint equation that relates accelerations.

\begin{align} \text{Constraint Equation: Velocities} \quad &\,\, \dot{L} = 0 = 2 \dot{L}_A + \dot{L}_B \\[5pt] \text{Constraint Equation: Accelerations} \quad &\,\, \ddot{L} = 0 = 2 \ddot{L}_A + \ddot{L}_B \end{align}

In these equations, it is important to remember that the values represent the changes in length, rather than direct measures of velocities. Though both vehicles would have positive velocities in the example above (velocities to the right), one \(\dot{L}\) value will be positive and one will be negative. This is because the truck is getting closer to the tree while the car is getting further away. A similar situation will occur for the accelerations, where both vehicles would have positive accelerations even if the \(\ddot{L}\) values are a mix of positive and negative values.

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/J1aPSrdDzdk.

Example \(\PageIndex{1}\)

A truck becomes stuck in the sand at a local beach. To help, a friend takes a rope 50 feet in length, ties one end to her car, loops the rope around a bar at the front of the truck, and then ties the other end to a stationary tree as shown below. If the car accelerates at a rate of 0.2 ft/s², what will the velocity of the truck be by the time it gets to the tree?

A pickup truck located 20 feet to the left of a tree is stuck in sand, and a car is 10 feet to the right of the tree. A rope has one end tied to the car's rear bumper, is looped through an attachment point on the front of the truck, and has its other end tied to the trunk of the tree. The rope is taut and horizontal throughout. — Figure \(\PageIndex{2}\): problem diagram for Example \(\PageIndex{1}\). A rope looped around the front of a truck, tied at one end to a tree trunk and at the other to a car's rear bumper, allows the car to pull the truck free of the sand it is stuck in.

Solution: Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/IxO_Nrs7Kj0.

Example \(\PageIndex{2}\)

A man has hooked up a pulley, a rope, and a platform as shown below to lift loads up onto a nearby rooftop. If \(x\) is currently 15 meters, \(y\) is currently 5 meters, and the man is walking away from the building at a rate of 0.5 meters per second, what is the current velocity of the platform?

A pulley is located on the roof at the right side of a building, 20 meters above the ground. A rope runs through the pulley, with one end tied to a load and hanging down the side of the building, and the other end held by a man on the ground to the right of the building. As the man walks further to the right, his distance from the building (x) and the height of the load above the ground (y) both increase. — Figure \(\PageIndex{3}\): problem diagram for Example \(\PageIndex{2}\). A man walks away from a 20-meter-high building, holding one end of a rope that passes over a pulley on the rooftop and raises a load on its other end.

Solution: Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/z3D_2jHLCik.