# Tension Question

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#### v4runc00l

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A 100kg man dangles a 50kg mass from the end of a rope. If he stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be? Answer is 333N.

I don't understand how they get this answer. I set up the following equation T - m(mass)*g = [ m(mass) + m(man)]a

I don't know where to go from there. Can someone help me? Thanks

#### v4runc00l

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how do we even know the mass is accelerating downward and that the man isn't providing enough force to keep the mass from moving?

#### illegallysmooth

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it's a frictionless surface.

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#### v4runc00l

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okay... so how do i set it up then?

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#### v4runc00l

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It is question 241 from Exam Crackers 1001 Physics. Essentially there's a cliff and a pulley out in open space right next to the edge and a mass dangling with a rope that is held by a man on the cliff.

#### TieuBachHo

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It is question 241 from Exam Crackers 1001 Physics. Essentially there's a cliff and a pulley out in open space right next to the edge and a mass dangling with a rope that is held by a man on the cliff.
I don't have that book but this is how you approach such problem (You can either wait on someone else or search through this web. Some one might post it here before). Set up the two objects with all forces acting on them independently. With T1 (tension on the man) and T2 (tension on the mass) are equal and so does the accelerations (a1 and a2) of both objects, you can find your T value solved. If you are not getting the answer, then check your misuse of +/- signs. I supposed this doesn't come with an angle, then it's more than just signs. Otherwise, I overthink yourself with a simple problem.

#### v4runc00l

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hmmm I still don't get the answer.. I think the trouble comes in because maybe I am not combining the forces in the x direction and forces in the y direction correctly?

#### BerkReviewTeach

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hmmm I still don't get the answer.. I think the trouble comes in because maybe I am not combining the forces in the x direction and forces in the y direction correctly?

In the y-direction, you have m1g acting downward on the object and T acting upward on the object. The net force on the downward object is m1g - T.

In the x-drection, you have the man accelerating across the frictionless surface, being impacted solely by T. His normal force is offset by his weight, so we need only consider his x-direction forces. Because the surface is frictionless, the only x-direction force acting on the man is tension.

Before we start the math, it's important to note that we have aman = aobject in magnitude, because they are connected by the rope.

In y: m1a = m1g - T
In x: m2a = T

Plugging m2a for T in the y-direction equation gives:

m1a = m1g - m2a

So,

m1a + m2a = m1g
(m1 + m2)a = m1g
a = m1/(m1 + m2) x g
a = [50/(100 + 50)] x g = 1/3 x g = 3.3

Plugging into the x-direction relationship gives:

T = m2a = m2(1/3g)
T = 100 x 3.3 = 333 N

This is an awful lot of math for one problem, so once you feel semi-comfortable with this answer, do the question again by POE. You never said what the answer choices are, but I'd be willing to bet you can eliminate two choices for sure if not three.

The object is falling, so T must be less than 500N (otherwise T would cancel the weight of the object and it wouldn't fall). T must have some value, otherwise the man and object would be accelerating at g.

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#### v4runc00l

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Thank you so much. Quick question though, how did you know the object was falling?

#### BerkReviewTeach

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Thank you so much. Quick question though, how did you know the object was falling?

Visualization.

If you had a rope connected to something dangling off a cliff and you were standing on ice with slippery shoes, you'd be pulled by the rope over the edge of the cliff. You could only stop sliding if you could plant your feet into the ice (spiked shoes = good friction).

The moral to the story is that you should never wear frictionless shoes on ice ponds near the edge of a cliff when holding on to ropes connected to falling masses that are hanging over the edge of the cliff.

Drawing a picture, as a few posters mentioned, can help with the visualization process.

#### v4runc00l

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Ahh I see. I have a picture here just couldn't understand why that was happening. Makes sense now. Thanks a lot!

#### BerkReviewTeach

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Ahh I see. I have a picture here just couldn't understand why that was happening. Makes sense now. Thanks a lot!

My pleasure. I just hope this helps prevent another slippery shoe on ice connected to a rope with a dangling mass over a cliff accident.

#### Longshanks

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Ahh I see. I have a picture here just couldn't understand why that was happening. Makes sense now. Thanks a lot!

My pleasure. I just hope this helps prevent another slippery shoe on ice connected to a rope with a dangling mass over a cliff accident.

Haha, yeah. As smooth pointed out, its a frictionless surface. So even if the man was somehow heavy enough to balance it out there's no static friction, so he'd have to get pulled down.

#### iceprincess13xx

##### New Member
How did you know to use the mass of the person instead of the mass of the block (or the mass of both combined)?

Thanks

-- Nevermind! Figured it out.

Removed
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#### GRod18

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what an great explanation by BR teach, but does this problem seem to be out of the MCAT league?

Answer choices btw are A. 250N B. 333N C.500N D.667N

#### jissho

##### Full Member
This problem caused me severe stress last night! Figured it out though.