Problem 6.32 Zach, whose mass is 77 kg , is in an elevator descending at 11 m/s
ID: 1331964 • Letter: P
Question
Problem 6.32
Zach, whose mass is 77 kg , is in an elevator descending at 11 m/s . The elevator takes 3.5 s to brake to a stop at the first floor.
Part A
What is Zach's weight before the elevator starts braking?
Express your answer to two significant figures and include the appropriate units.
Part B
What is Zach's weight while the elevator is braking?
Express your answer to two significant figures and include the appropriate units.
Problem 6.33
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 kg , and the doctor has decided to hang a m = 6.8 kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed. (Figure 1)
Part A
Determine the amount of tension in the rope by using Newton’s laws to analyze the hanging mass. Assume that = 17 .
Hint: If the pulleys are frictionless, which we will assume, the tension in the rope is constant from one end to the other.
Express your answer to two significant figures and include the appropriate units.
Part B
The net traction force needs to pull straight out on the leg. What is the proper angle for the upper rope?
Express your answer to two significant figures and include the appropriate units.
Part C
What is the net traction force pulling on the leg?
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Problem 6.32
Before the elevator starts braking:
W=ma (because there is no acceleration, a is just acceleration due to gravity, 9.81ms^-2)
W=(77kg)(9.81ms^-2)
= 755.37 N
When the elevator is braking:
By dividing 11m/s by 3.5s, I find the elevator's deceleration, which is 3.14ms^-2.
Since this causes an increase in the normal force, then it is:
W=ma (where a this time equals 9.81+3.14)
=(77kg)(12.95ms^-2)
= 997.15N
Please post the other question separately