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
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
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?
Well, it seems the boy has found himself in quite a "tire-some" situation, eh? Now, let's put our thinking caps on and figure out the answer!
The force with which the boy must pull on the free end of the rope to support his weight in the tire can be determined by considering the forces at play. We have the weight of the boy plus the tire, which is W, acting downwards. This force must be balanced out by an equal and opposite force in the upward direction.
Now, due to the rope passing over the pulley, the force required to support his weight is effectively halved. So, the force with which the boy must pull on the free end of the rope is (1/2)W.
Therefore, the answer is (a) (1/2)W. The boy doesn't need to pull too hard, just enough to keep himself "tire-ly" balanced! Keep those questions coming, my friend!
To find the force with which the boy must pull on the free end of the rope to support his weight in the tire, we need to consider the balance of forces acting on the system.
The force of gravity acting on the boy and the tire is equal to the weight of the system, which is given as W. This force is pulling the boy and the tire downward.
To counteract the force of gravity, the boy needs to apply an equal and opposite force on the free end of the rope. This force will be transmitted through the rope, passed over the pulley, and back down to the boy's hands.
In an ideal situation, without any friction or other external forces, this pulling force would be equal to the weight of the system, which is W. Therefore, the correct answer is (b) W.
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.