There Is A 5Kg Block Resting On A Table, This Block Is Attached To A Rope That Goes Through A 10Kg Wheel Attached To The Edge Of The Table.

[sandra123123]

There is a 5kg block resting on a table, this block is attached to a rope that goes through a 10kg wheel attached to the edge of the table. At the end of the rope is a 2.5kg brick dangling 1 meter off the wheel/table, The coefficient of friction between the block and the table is 0.4. If the system is initially at rest, how long will it take for the 2.5kg brick to drop 1m to the floor after it is released? (Assume the friction on the wheel's axle is negligible and there is no slippage between the rope and the wheel.)

[Tilly2]

This can't be homework because schools have broken up. I hope someone will answer this for you soon, sandra. I can't.

[vascop]

What do you mean exactly when you say the rope goes "through" the wheel? Is it being used like a pulley, so do you mean that the rope goes round the rim of the wheel? Why do we need the weight of the wheel if it's attached to the table?

[Peaspeculiars]

The 2.5kg brick does not move at all. The downwards force has to be greater than that exerted by the counter weight of 5kg on the table, but as it isn't, it will stay put.

I think so anyway...

[vascop]

No, no, no, the rope is not pulling against the weight of the block because that acts downwards, the rope is pulling against the frictional force which is 5 x g x 0.4 =2.0 x g.

[bhg481]

The frictional force trying to hold the block in place is 0.4 x 5 = 2Kg.

The force trying to move the block via the pulley is 2.5Kg.

Using Force = mass x acceleration you can calculate the acceleration of the block.

Knowing the acceleration using t=sqrt (2 x 1metre/acceleration) gives the time for the brick to fall 1 meter.

[Peaspeculiars]

Doh! reading the question, I thought it was a lifting force as the pulley was raised, never mind, bottom of the class again ;o) sorry Sandra, ignore my previous answer

[sandyRoe]

what relevance has the weight of the 10kg wheel?

[jim360]

As vascop says, the fact that the block on the table is heavier makes no immediate difference, since its weight is acting entirely into the table. If the table were perfectly smooth the only force acting in the system would be the weight of the dangling brick.

The way to approach this is as follows:

Step 1: draw free-body diagrams for the two blocks and the wheel. At this point it makes sense to be as general as possible, so call the large mass M and the other mass M, with a coefficient of friction mu, and acceleration to gravity g.

Step 2: Resolve forces in the horizontal and vertical directions for all objects.

Step 3: Use negligible friction/ no slippage to assume that the tension in the rope is equal everywhere.

Step 4: Input above results into Newton's second law, for both blocks, in the directions they can move.

Step 5: Inextensible string will imply that the accelerations of both blocks are the same. If you've done all the working above correctly this will turn your problem into a pair of simultaneous equations in the Tension of the rope, T, and the acceleration a.

Step 6: Eliminate T from these equations, so that you are left with an equations that gives acceleration in terms of the masses, friction coefficient, and gravity.

Step 7: Input the numbers for this problem into the equation, and then you will no what the constant acceleration of the brick is.

Step 8: Use the SUVAT equation s= ut + (1/2)at^2 and solve this for t, with u = initial velocity = zero.

Following the above steps should lead you to the answer.

[jim360]

I've not bothered to work out precise numbers but the method above leads me to the conclusion that the block and brick won't move if the coefficient of friction is 0.5 or greater.

[jim360]

The way I see it, the brick is dangling over the edge of the table so only ever falls straight down. It's at a point like this that a diagram would help to clarify the picture exactly.