Calculate the mass of sand added to the bucket

[WebassignHelp]

Please Help!

I cannot figure out this problem. Please help out there. Thank you! I don't know where to start.


A 25.5 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is 0.450 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.

(a) Calculate the mass of sand added to the bucket.
?kg

(b) Calculate the acceleration of the system.
?m/s2 (downward)

 

[cristo]

Welcome to the forums. Please note that you must show some work before we can help you. Have you not studied any of this material in class, or have any relevant class notes on the topic from which you can begin?

 

[WebassignHelp]

I have, but the mass of the bucket would normally be given to me and I cannot figure out how to solve for the mass that needs to be applied in order for it to move. Taking the 1.00kg of the bucket into consideration, of course. Once I have the mass I can then solve for acceleration.

 

[cristo]

Can you calculate the static frictional acting on the block? Can you calculate the force of the string on the block? What can you say about when the block begins to move?

 

[WebassignHelp]

The fiction force should be maxed out when the bucket begins to lower.

 

[Physwiz]

here's your hint:

As of this scenario, the system remains static as long as the forces on the block(due to friction)...and on the bucket+sand(gravity), balance each other!

let's say 'x' is the maximum mass of sand for which the system remains stationary...then equating the forces

force on block(use the static coefficient)=force on bucket+sand

Also,
force on bucket+sand=(1+x)*g

 

[cristo]

That's not a hint, Physwiz, that's a recipe for the solution! If the OP had answered my questions, he would have been able to obtain something like that.

 

[Physwiz]

It's still not the answer.