Damn computer iq tests

[Peter Pan]

I took one of those stupid computer IQ tests and one of the questions is still nagging my poor tired brain.

a monkey and a weight are on the opposite sides of a pully. The rope, that the monkey is holding on to and the weight is tied to, is considered weightless. The pully is also fricitonless. The monkey and the weight are the same weight.

Q) If the weight and the monkey are both hanging on the rope at the same distance from the pully and are at an equalibrium, what happends to the weight if the monkey climbs up the rope(towrds the pully)

A) the weight rises
2) the weight lowers
C) the weight stays in the same spot

I said c. if the rope is weightless, it shouldn't matter where the opposites weights are on the rope. For some reason I cannot convince myself that I am correct. What do you think?

 

[revesz]

The weight of the monkey will not change(because he dosn't gain mass), so the rope will pull down on the pully with the same strength. The direction that the rope is pulling will not change.

So the C is the correct answer.

 

[joshi-wan kenobi]

Does that mean if we were the monkey we could sit there yanking on the rope and the only thing that would happen is we would move up? Maybe it's just me and my lack of monkey experience but somehow that doesn't sound quite right.

It seems like the weight and your monkey self would rise. lol In any case, I can see why this bugged you so much, Peter. :D

 

[Moose352]

So where does the monkey climbing force force go?

 

[revesz]

oh, I see what the question is asking. When the monkey first starts to climb up, he will weigh more, after the monkey stops accelerating his weight will return to normal. So there will be a moment when the monkey weighs a little more. If the monkey decides to stop climbing the rope, then he will weigh less, until he has finished his de-acceleration, at which point his weight will return to normal. Think of an elevator, when going up you weigh a little more during the acceleration, then you feel normal, and when you reach your floor you lose a little weight for a few seconds as the elevator slows.

So then my first answer would still be correct.

Unless you say that the monkey is always climbing a little bit faster each second, then it's(or his/her) weight would be increased. I would interpret that the question is not asking that so your answer should be C.

 

[krab]

A. Weight rises

 

[FUNKER]

throught the equation torque=Fr if the monkey moves closer to the pulley he has less radius therefore less torque and inturn the wieght will lower because it will have more torque than the monkey.

if u get 124 in an IQ tests tht good

 

[scientist0]

yep

i think its A.


until the weights get up to the same height...to balance?rite

 

[revesz]

I still think that the answer is c. Can you tell me where this IQ test was and I'll take to see who's the smartest.

The ONLY reason the monkey would change in weight is because he would be farther away from the source of gravity, so he/she(the monkey) would weigh slightly less.

 

[HallsofIvy]

throught the equation torque=Fr if the monkey moves closer to the pulley he has less radius therefore less torque and inturn the wieght will lower because it will have more torque than the monkey.

No, torque has nothing to do with this. The "radius" is the distance from the axis of rotation to a point perpendicular to the applied force and that is NOT the distance to the monkey. The radius would be the radius of the pulley itself which does not change and is assumed to be small compared with the other distances.

The force with which the monkey pulls on the rope, in order to pull himself up the rope, is exactly the force the rope applies to the weight. Since the monkey and the weight have the SAME mass, the weight rises at exactly the same rate the monkey does.

 

[Tyro]

The issue is whether the tension in the rope remains the same when the monkey is static and when the monkey is climbing up.

If the tension is the same, then the weight is in equilibrium and doesn't move. If the tension increases, the weight has to accelerate upwards. If you consider a free body diagram of the monkey, for it to climb upwards, it has to accelerate upwards first, so it has to exert more force than its weight (or the counterweight).

So the answer is A as the tension in the rope increases, the weights will rise. C is only true if the monkey's acceleration is infinitesimal. The case when the counterweight does not move (C) would be if the problem was symmetrical, i.e. you also had an equivalent monkey on the other side pulling itself upwards just as hard.

BTW, Peter Pan, you labelled the answers A,2,C instead of A,B,C

 

[turin]

There are two coupled free-body diagrams: the monkey and the weight.

Initially, W_monkey = T, W_weight = T,
where T is the tension in the rope.
v_monkey = 0, v_weight = 0
W_monkey = W_weight = W.

Assume that T is always instantaneously uniform throughout the rope between the monkey and the weight no matter what happens (in other words: Y -> infinity, no elasticity, the rope does not stretch).

There is a change of state by the monkey acquiring an upwards velocity. After the change of state, the equations:
W_monkey = T, W_weight = T,
still hold.
W_monkey = W_weight = W.

The monkey received an impulse, Δp, from the rope. Newton's third law says that the rope received the negative impulse from the monkey, -Δp.

Since T = W_weight, then the weight receives this negative impulse. In other words, the weight gains the same momentum towards the pulley, Δp.

Since W_monkey = W_weight, the weight gains the same velocity towards the pulley as the monkey.

The answer is a.

 

[Soveraign]

I vote D) "poorly worded question" and also I give it an F for having to do more with knowledge than a measure of intelligence.

It's A) if you are considering the moments when the monkey is accelerating to a climbing speed. It's C) if you are considering when the monkey is climbing at constant velocity, and for that matter it's B) if you technically take into account the freakin' monkey is getting further away from the center of the gravitational force (earth) and thus weighs less.

 

[Moose352]

I absolutely agree with ^.

 

[TonySlim]

I don't think the elevator analogy is a good one, at least not to deal with the energy going into the climbing. But if it is a good one, wouldn't the rising weight (per answer "A") also get heavier? As heavy as the heavier monkey? And cancel out the rising...but then it wouldn't be heavier anymore...then it would rise...

 

[revesz]

Sorry, I was wrong, I wasn't reading the question right.

So now I have changed my answer to A),.

hey, anyone know where a good IQ test is, I've looked around the net and my scores have been really inconsistent.

 

[Tyro]

My experience with online IQ tests is that they always give you an IQ in the genius range, simply because they want you to buy their premium tests/result analyses. And visit their site more often to rub your ego as well. Nice marketing ploy, though. If you want a real test, get it done IRL under actual test conditions (no sneaky calculator multitasking ) by a recognised institution or agency.