Crazy Pulley Problem: Solve for Distance, Acceleration & Tension

[spursfan2110]

Homework Statement

Part 1:

A 17 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 5.9 kg block via the arrangement shown (attached below). The acceleration of gravity is 9.81 m/s/s. Find the horizontal distance the 17 kg block
moves when the 5.9 kg block descends a distance of 9.3 cm. The pulleys are massless and
frictionless. Answer in units of cm.

Part 2:

Find the acceleration of the 17 kg block. Answer in units of m/s/s.

Part 3:

Find the tension in the connecting string. Answer in units of N.

Homework Equations

F = ma

The Attempt at a Solution

I am absolutely stumped on all three...I don't want someone to solve this for me, but any little hint to get me started would be great. My main problem is that none of the equations I have have anything to do with distance and so I don't know how to relate that to these forces and acceleration.

 

[Spinnor]

Draw a free-body diagram of the two masses of interest. Note that there is a relationship between accelerations of the two masses, I think one is twice the other.

 

[spursfan2110]

Ok, I've drawn the diagram for the hanging mass, and with T being an upward force and mg being a downward force, ma being mg -T, but I am really confused about the mass on the table. Gravity and the normal force cancel each other out obviously, and are there two tension forces acting on it as well?

 

[rl.bhat]

Since single string is connecting the two masses, the tension is each segment of the string must be equal.
So m2*a = ...? Here m2 = 17 kg

 

[sArGe99]

Since the first pulley(the smaller one) is physically connected to the first block, you have to treat the pulley as being a part of the block. So the force experienced by the (block+small pulley) is 2T while the tension force in the other block is only T. Remember that tension in the string is the same throughout since all pulleys are said to be massless and frictionless.
2T = 17 a
and
5.9*g - T = 5.9*a