See the diagram here:

This was a question on my son’s 11th grade physics test. I’m assuming the 11.3N and 5.3N blocks are weights on the ropes.

Any help?

See the diagram here:

This was a question on my son’s 11th grade physics test. I’m assuming the 11.3N and 5.3N blocks are weights on the ropes.

Any help?

Looks like a mistake to me.

N - Newton, which is a Force. Force implies a direction, which isn’t indicated.

The Mass at the bottom, 18.6kg is correctly labeled.

So, either the vectors of the forces are missing, or they are actually masses, and should be in kg, not N.

I wondered about that, too. My son told me those are weights, so I’m assuming they are 11.2/9.8 = 1.143kg and 5.3/9.8 = 0.54kg, respectively.

I figured that was just the instructor’s way of getting the students to have to think about the physics in force vs. mass and do a conversion.

Well a Newton is 1 (Kg *M)/S^2 so the downward force from the 18.6 Kg weight is:

18.6 Kg * 9.8M/S^2 = 182.29 N

Figure the upward forces have to equal the downward force…

Hmm…

Just took another look at that diagram - it is kind of goofy.

My son tried to solve it by using:

18.6kg * 9.8m/s^2 = 182.28N

T/2 + T/2 - 16.5N = 182.28N

T = 182.28N + 16.5N = 198.78N

He was given 4/5 points on the question.

Disregarding the distances of he ropes from the center of the mass of the 18.6 bar, this is what I would have done.

I assumed that the 11.2N and the 5.3N represented the tension in the cables (tension is a force, or a stress, depending on how you specify it). You don’t need a direction, since cables can only support tensile forces, not compressive forces.

That still doesn’t lead to a solution, but at least it makes the question make sense.

From what is drawn and written, it doesn’t seem like the problem isn’t clearly stated. Does your son know that what he thought the problem described is the same as what the instructor thought the problem described?

Cables (and ropes, strings, wires, chains,belts, etc ) are one tension , or the only way they can be at different tension is from friction.

Thats some massive friction we are contending with, perhaps the system is in a hurricane and the wind shear is 150 miles per hour and its on a very high gravity planet.

Ignore the tension/weight blocks, they forgot to the delete them !.

Then you can answer the question.

The weight is being held by 4 strings, equally, and the pulley in question has two strings. the answer is half the weight (mass * force of gravity downward), but upward ( equal and opposite, and all that.)

Not necessarily half. That would only be true if the situation were symmetrical. Since dimensions are not supplied, we don’t know if that is the case.

Assuming that the newton values are forces in the rope and not weights, the answer should be 111.86 N (approximately, I rounded the weight up to 182.5 N).

You just label the different forces and then solve a system of linear equations.

http://awwapp.com/s/04/a4/48.html

T1+T2=182.5

T1=T3+11.2

T2=T4+5.3

etc.

Am I missing something?

A weight is being suspended by two cables.

Why would the cables have different tensions?

It matters not how many pulleys they go through.

Seems the left pulley is not attached to the ceiling so the whole thing would fall apart anyway.

Seems too obvious so what am I missing?

As a physics teacher, I have to say that is a very poorly designed problem.

Er, yeah, and assuming frictionless pulleys and that the blocks are weights as stated in the OP, I get the tension is ~100.5.

Equal tension lengths are as shown in newme’s drawing. There are 7 unknowns, make seven equations and solve. 11th graders should have that much linear algebra.

My answer, referring to this drawing: http://awwapp.com/s/a9/64/ea.html

T = 2*T1

T2=T1-11.2 (Newtons)

T3=T2-5.3=(T1-11.2)-5.3=T1-16.5 (Newtons)

WEIGHT OF BAR = 18.6kg*9.81 m/s = 182.466 (Newtons)

T1 + 2*T2 + T3 = 182.466
–>T1 + 2*(T1-11.2) + (T1-16.5) = 182.466

–>4

–>T1= (182.466+38.9)/4 = 55.3415

T = 2

ETA: If T1 and T3 are unequal, wouldn’t this cause a torque in the bar about its center? So the bar would rotate clockwise and hit the pulleys.

I set up my equations incorrectly here. I agree with the 110.7.

Regarding rotation: it depends on where the cables are attached with respect to the blocks center of mass.

Any chance that the two blocks are springs with the given tension?

I have found that when someone goes to extraordinary trouble to ask for help with an exam question (like you did), it usually means they have tried every possible source for help they could and are now stumped. What that usually means is there was a mistake made by the person preparing the exam. Not always. But usually.

I didn’t look at the question because I’m hoping that someone has indeed found there was a mistake made in the preparation of the exam and this thread was then closed. If not, I would hope that someone else would verify my experience that usually this means a mistake was made with the preparartion of the exam question.

In any event, I wish you the best of luck.

I should also mention that on more than one occasion (although not many more), one of my high school teachers (who was an alcoholic) either taught a class while drunk or prepared a test while drunk (apparently) and one or more of the test questions were just bonkers and had to be excused. (Oh, boy! Was his face red!)