An accident victim with a broken leg is being placed in

Chapter 6, Problem 33P

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QUESTION:

II An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of \(4.0 \mathrm{~kg}\), and the doctor has decided to hang a \(6.0 \mathrm{~kg}\) mass from the rope. The boot is held suspended by the ropes, as shown in FIGURE P6.33, and does not touch the bed.

a. Determine the amount of tension in the rope by using Newton's laws to analyze the hanging mass.

b. The net traction force needs

to pull straight out on the leg. What is the proper angle \(\theta\) for the upper rope?

c. What is the net traction force pulling on the leg?

                                         

Equation Transcription:

Text Transcription:

4.0 kg

6.0 kg

theta

Questions & Answers

QUESTION:

II An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of \(4.0 \mathrm{~kg}\), and the doctor has decided to hang a \(6.0 \mathrm{~kg}\) mass from the rope. The boot is held suspended by the ropes, as shown in FIGURE P6.33, and does not touch the bed.

a. Determine the amount of tension in the rope by using Newton's laws to analyze the hanging mass.

b. The net traction force needs

to pull straight out on the leg. What is the proper angle \(\theta\) for the upper rope?

c. What is the net traction force pulling on the leg?

                                         

Equation Transcription:

Text Transcription:

4.0 kg

6.0 kg

theta

ANSWER:

Step 1 of 3

a.)

We have to determine the amount of tension in the rope by using Newton's laws to analyze the hanging mass.

The free body diagram for the hanging mass is as shown in the figure below.

                                                 

The Tension \(\vec{T}\) in the rope can be found from the above free body diagram.

                                             \(T=F_{G}\)

                                            \(T=m g\)

Where,

\(m=\) hanging mass \(=6.0 \mathrm{~kg}\)

\(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\)

Thus,

                                           \(T=(6.0)(9.80)\)

                                          \(=58.8 \mathrm{~N}\)

Therefore, the amount of tension in the rope is \(58.8 \mathrm{~N}\).

b.)

 

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