## Mastering Physics Solutions: Tension in a Massless Rope

Tension in a Massless Rope

Part A = T1
Part B = T2
Part C = 3rd
Part D = equal to
Part E = Fu = Fd
Part F =

• The tension in the rope is everywhere the same.
• The magnitudes of the forces exerted on the two objects by the rope are the same.
• The forces exerted on the two objects by the rope must be in opposite directions.
• The forces exerted on the two objects by the rope must be in the direction of the rope.

Solutions Below:

Consider the three sections of rope labeled a, b, and c in the figure.

• At point 1, a downward force of magnitude Fad acts on section a.
• At point 1, an upward force of magnitude Fbu acts on section b.
• At point 1, the tension in the rope is T1.
• At point 2, a downward force of magnitude Fbd acts on section b.
• At point 2, an upward force of magnitude Fcu acts on section c.
• At point 2, the tension in the rope is T2.

Assume, too, that the rope is at equilibrium.

Part A

What is the magnitude Fad of the downward force on section a? Express your answer in terms of the tension T1.

We know from the given information that at point 1 downward force of magnitude Fad acts on section a and that the tension at point 1 is T1. Since the two are the same thing, we know that:

Part B

What is the magnitude Fbu of the upward force on section b? Express your answer in terms of the tension T1. We know that the downward force (Part A) is the same as the upwards force – so our answer remains the same:

Fbu =T1

Part C

The magnitude of the upward force on c, Fcu, and the magnitude of the downward force on b, Fbd, are equal because of which of Newton’s laws?

Recall Newton’s 3rd law: To every action there is always an equal and opposite reaction. So the answer is:

Newton’s 3rd law

Part D

The magnitude of the force Fbu is ____ Fbd.

It is important to realize that Fbu and Fbd are not a Newton’s third law pair of forces. Instead, these forces are equal and opposite due to the fact that the rope is stationary (ab=0) and massless(mb=0). By applying Newtons first or second law to this segment of rope you obtain:

Fb net = Fbu – Fbd = mbab = 0, since mb = 0 and ab = 0. Note that if the rope were accelerating, these forces would still be equal and opposite because mb = 0.

equal to

Part E

Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces?

The forces on the two ends of an ideal, massless rope are always equal in magnitude. Furthermore, the magnitude of these forces is equal to the tension in the rope.

Therefore:

Fu = Fd

Part F

A massless rope is attached at its ends to two stationary objects (e.g., two trees or two cars). For this situation, which of the following statements are true?

All of the below are true:

• The tension in the rope is everywhere the same.
• The magnitudes of the forces exerted on the two objects by the rope are the same.
• The forces exerted on the two objects by the rope must be in opposite directions.
• The forces exerted on the two objects by the rope must be in the direction of the rope.

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