3rd Law: Weight vs Tension (picture)

1. Nov 10, 2009

merzperson

1. In case a in the figure block A is accelerated across a frictionless table by a hanging a 10N weight (1.02kg). In case b, block A is accelerated across a frictionless table by a steady 10N tension in the string. The string is massless, and the pulley is massless and frictionless.

Is A's acceleration in case b greater than, less than, or equal to its acceleration in case a?

3. I just don't see how a 10N weight differs from a constant 10N downward force on the rope. How is it that the acceleration of block A in case b is greater than in case a? I assume there is some kind of difference because in case a we are comparing two objects and in case b we just calculate the net force on block A, but I can't figure out how to do this.

2. Nov 10, 2009

rl.bhat

If T is the tension in the rope, in the first case T> 10 N. Where as in the second case T = 10 N. There fore the acceleration in case a is smaller than b.

3. Nov 10, 2009

merzperson

Your reply implies that as the Tension of the rope increases, the acceleration of block A decreases. This is the exact opposite of what I would expect. Explain?

Also, could you please explain to me how the Tension in case a is greater than 10N?

Thank you.

4. Nov 10, 2009

rl.bhat

In the first case
mg - T = 10N/g*a, where a is the common acceleration hanging weight and block A.
Similarly
T = mA*a.
In the second case T itself is 10 N.

5. Nov 10, 2009

merzperson

Thank you very much for your replies. However, I am still having trouble.

I don't understand where you got the equation:
mg - T = 10N/g*a

Or for that matter the other equation. Is mA the mass of block A?

Now I do believe I understand why the tension in case a is not 10N; since the blocks are in motion, the 10N block will not have the effect of 10N. However, I do not know how to tell whether the tension will be greater than or less than 10N, and I haven't a clue how to calculate the tension. Any help with this would be greatly appreciated.

6. Nov 10, 2009

rl.bhat

Acceleration of the hanging mass is less than g, because it is not falling freely.
So its acceleration is given by the equation
mg - T = ma ------(1), where m is the mass of the hanging block.
Acceleration of the block is given by
T = mA*a ----(2), where mA is the mass ob block A.
Hanging block and block A must move with the same acceleration, because they are connected by a single rope. The common acceleration can be found by solving eq.1 and 2. It comes out to be
a = mg/(m + mA). = 10N/(m + mA)
In the second case acceleration of the block A is
a' = 10N/mA. Now compare the accelerations.