Friction Pulley System

Sep 2013
10
0
hi i just want to ask how to solve a pulley system with a box and in that box there is also a box (pic) https://fbcdn-photos-e-a.akamaihd.net/hphotos-
ak-prn2/1235923_673131749382598_
1392289205_a.jpg


do u jist add the box a and box b with the 10kg?

i only know how to solve a simple pulley system but idk how to solve it if there iis a box in the top of the box. pls help me thank u

btw if u don't get it and u want to solve it here's the given
box a has 45kg
box b has 50kg
acceleration is 0.63m/s upward and to the left

but my ques is only how do i solve if there is a box on the top of the box thankd
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
It depends on the coefficient of friction between the 10Kg weight and Box A. If you assume that friction=0, you can ignore the 10Kg weight. But if the coefficient of friction 'u' is greater than zero then it gets more complicated. If 'u' is greater than 1 the weight will move with Box A and not slide relative to it, so the weight is added to Box A for yuor calclations. If 'u' is less than 1 the weight's maximum acceleration can be a=ug, so if the system acelerates at less than that the weight stays on Box A, and if acceleration is greater than that the weight accelerates at ug and hence slides backwards relative to Box A.

I don't understand this statement: "acceleration is 0.63m/s upward and to the left" - please clarify this, because from the drawing it's apparent that the acceleration of box A is to the right, and would be at least 4.66 m/s^2.
 
Sep 2013
10
0
It depends on the coefficient of friction between the 10Kg weight and Box A. If you assume that friction=0, you can ignore the 10Kg weight. But if the coefficient of friction 'u' is greater than zero then it gets more complicated. If 'u' is greater than 1 the weight will move with Box A and not slide relative to it, so the weight is added to Box A for yuor calclations. If 'u' is less than 1 the weight's maximum acceleration can be a=ug, so if the system acelerates at less than that the weight stays on Box A, and if acceleration is greater than that the weight accelerates at ug and hence slides backwards relative to Box A.

I don't understand this statement: "acceleration is 0.63m/s upward and to the left" - please clarify this, because from the drawing it's apparent that the acceleration of box A is to the right, and would be at least 4.66 m/s^2.


hi sorry. ok so this is thd real picpls help me thanks! i don't really get it where to put the 15kg
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Your lipstick drawing isn't coming across very well. I don't understand the 0.63 m/s^2 bit - what's making that occur? Nor do I understand why this figure is so inconsistent with the original figure you posted. We'd love to help, but you need to do a better job at helping us understand the problem first.
 
Sep 2013
10
0
Your lipstick drawing isn't coming across very well. I don't understand the 0.63 m/s^2 bit - what's making that occur? Nor do I understand why this figure is so inconsistent with the original figure you posted. We'd love to help, but you need to do a better job at helping us understand the problem first.
I'm really sorry coz i am just usingmymobile todraw the figure anyways hereisthe clearer version and nevermind the original figure thank u so much. I know how to solve this but ican't do if there is a box on thetop of abox also thank u
 

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topsquark

Forum Staff
Apr 2008
2,976
630
On the dance floor, baby!
Is the top box able to slide on the bottom box? If so you'll have to add an expression for friction to your solution.

-Dan
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Finding T is straight forward. The friction force fk depends on the coefficient of friction between box A and the table - did they give that value to you? The normal force that contributes to fk would include the weight of the 15Kg mass. As for the free body diagram for Box A: if you assume zero friction between Box A and the 15Kg weight then the weight doesn't accelerate and you have F-fk-T=Ma, where M = mass of box A. On the other hand if the coefficient of friction is at least equal to a/g then the 15Kg will accelerate with Box A and M is the sum the two. If u is greater than 0 but less than a/g then the weight has a free body diagram of umg=ma, so it accelerates at a=ug. and the forces acting on Box A are F-fk-T-umg, where m=15Kg. Hope this helps.
 
Sep 2013
10
0
Finding T is straight forward. The friction force fk depends on the coefficient of friction between box A and the table - did they give that value to you? The normal force that contributes to fk would include the weight of the 15Kg mass. As for the free body diagram for Box A: if you assume zero friction between Box A and the 15Kg weight then the weight doesn't accelerate and you have F-fk-T=Ma, where M = mass of box A. On the other hand if the coefficient of friction is at least equal to a/g then the 15Kg will accelerate with Box A and M is the sum the two. If u is greater than 0 but less than a/g then the weight has a free body diagram of umg=ma, so it accelerates at a=ug. and the forces acting on Box A are F-fk-T-umg, where m=15Kg. Hope this helps.
i already know the Tension but I can't solve for fk bcoz there is no coefficient given. do u know how to solve for fk even if there is no coeeficient of friction given?
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
i already know the Tension but I can't solve for fk bcoz there is no coefficient given. do u know how to solve for fk even if there is no coeeficient of friction given?
To determine the coefficient of friction you would have to be given the value for F as well as a.
 
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ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Are you sure you have the complete problem? Is it from a text book? Perhaps they want the value of F+fk?