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Hanging Object Tension

 
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Oct6-07, 01:12 PM   #1
 

Hanging Object Tension


1. The problem statement, all variables and given/known data

A light rope is attached to a block with mass 3.70kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.9N.

What is the acceleration of either block?
Find the mass m of the hanging block.
How does the tension compare to the weight of the hanging block?


2. Relevant equations

F=ma
w=ma

3. The attempt at a solution

After reading the textbook (University Physics), I was able to determine that in a similar problem, that the had these two equations.

T =m1a
m2g -T = m2a

They eliminated T and getting, m2g=m1a+m2a = (m1+m2)a.

In this problem the tension is given.

So.

18.9 = 3.7 (a)
m2(9.8) - 18.9 = m2(a)

we want a, so we need to solve for it.

Thus far we got that
18.9/3.7 = a = 5.11.


now we want mass, we plug it in into the second equation.
9.8m_2 - 18.9 = 5.11m_2
-18.9 = -4.69m_2
m_2 = 4.03 kg.

Is this how on should solve it. I already drew the free body diagram in which for the 3.70kg object there are three forces, normal and weight which are equally the same and tension to the right which is a bigger force. Hence it is moving the right.

For the object hanging there is smaller tension force, and larger weight force, this is because the object on the horizontal is moving to the right, and thus means that there must be enough weight hanging for it to move.

but now, if all is right above (including logic), then how do I calculate the following..

How does the tension compare to the weight of the hanging block? :(
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Oct6-07, 01:26 PM   #2
 
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everything looks right. just calculate the weight of the hanging block... you have the mass already... where are you getting stuck?
Oct6-07, 01:30 PM   #3
 
got it, I figured just checking to see if all my previous work was correct. :)thank you learningphysics.
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