Well, downward forces are W, and upward forces are his Pull, and tension.
So his pull + Tension=W on the tire side.
On the other side, tension=Pulling force
so
Pull+tension=W
Pull+ Pull=W
do the math.
A boy sits in a tire that is attached to a rope that passes over a pulley fastened to the ceiling and then passes back down to the boy's hands. The weight of the boy plus the tire is W. The force with which the boy must pull on the free end of the rope to support his weight in the tire is:
(a) (1/2)W
(b) W
(c) 2W
(d) (2/3)W
(e) (3/2)W
3 answers
Ok, I think I got it. Are these free body diagrams correct? (ignore the dots)
TIRE+BOY:
T F
|.|
.|
W
FREE END:
.T
.|
.|
.F
SYSTEM:
.F
.|
.|
.T.....(This tension is down)
T.F....(This tension is up)
|.|
.|
.W
CANCELING TENSIONS:
2F
|
|
W
2F=W so F=(1/2)W
TIRE+BOY:
T F
|.|
.|
W
FREE END:
.T
.|
.|
.F
SYSTEM:
.F
.|
.|
.T.....(This tension is down)
T.F....(This tension is up)
|.|
.|
.W
CANCELING TENSIONS:
2F
|
|
W
2F=W so F=(1/2)W
Now I'm confused. Why does changing the direction of a force double it? I'm also confused about whether the tension is a reaction to the weight of the tire and the boy or the force that the boy pulls down on the rope with. Also if the force in the free body diagram of the system were another person pulling the tire and the boy up, why would the force not be half the weight in that case?
2 answers