The figure shows two M B =1.00 kg blocks connected by a rope. A second rope hang

Question

The figure shows two MB=1.00 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of MR= 250 g . The entire assembly is accelerated upward at a= 3.00 m/s2 by force F? .(Figure 1) This question will take you through both symbolic and numerical calculations for the forces on the blocks and ropes. Take up to be the +y direction.

Determine the magnitude and direction (+ or ?) of the vector F.

Give your answer in terms of some or all of the variables MR, MB,a, and gravitational acceleration g.

Determine the magnitude and direction (+ or ?) of the tension at the top end of rope 1.

Give your answer in terms of some or all of the variables MR, MB,a, F, and gravitational acceleration g.

Determine the magnitude and direction (+ or ?) of the tension at the bottom end of rope 1.

Give your answer in terms of some or all of the variables MR, MB,a, F, and gravitational acceleration g.

What is the tension at the top end of rope 2?

Give your answer in terms of some or all of the variables MR, MB,a, F, and gravitational acceleration g.

I have all of the actual numerical values, I just need the equations with all of the variables that Mastering Physics requires

Explanation / Answer

To begin working this out, first of all you use F = ma with the entire system's mass and F as the accelerating force. This assumes that all other forces cancel out.

a)

FOR ENTIRE SYSTEM:

F = (1+1+0.25+0.25)*3
= 2.5*3
= 7.5N


b)

Now you take into account the individual blocks, still keeping in mind each is accelerating at 3ms^-2. Block A has an acceleration of 3, a mass of 1kg, and is also being pulled *in the opposite direction* by the tension of rope 1.

FOR BLOCK A ONLY:

F - T = overall accelerating force.

Using F = ma:

(F - T) = m*a
7.5 - T = 1 * 3
T = 7.5/3
T = 2.5N.

c)

Since we know that the rope is accelerating, now we can apply the next step of the problem. We use F = ma for the rope.

FOR ROPE ONE ONLY:

Tension at top of rope1 - tension at bottom of rope1 = overall accelerating force.

2.5 - T = ma
2.5 - T = 0.25 * 3
2.5 - 0.75 = T
T = 1.75N.

d)

Tension at bottom of rope1 - Tension at top of rope2 = overall accelerating force

1.75 - T = 1 * 3

T = 1.75 - 3

T = -1.25 N

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