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Sep 10th 2013, 05:12 AM

#1  Junior Member
Join Date: Sep 2013
Posts: 10
 Friction Pulley System
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) [IMG]https://fbcdnphotosea.akamaihd.net/hphotos
akprn2/1235923_673131749382598_
1392289205_a.jpg[/IMG]
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

 
Sep 10th 2013, 09:28 AM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,347

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 10th 2013, 05:33 PM

#3  Junior Member
Join Date: Sep 2013
Posts: 10

Originally Posted by ChipB 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 
 
Sep 10th 2013, 07:58 PM

#4  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,347

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 10th 2013, 08:43 PM

#5  Junior Member
Join Date: Sep 2013
Posts: 10

Originally Posted by ChipB 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

 
Sep 11th 2013, 04:09 AM

#6  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,730

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
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Sep 11th 2013, 04:31 AM

#7  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,347

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 FfkT=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 FfkTumg, where m=15Kg. Hope this helps.

 
Sep 11th 2013, 05:28 AM

#8  Junior Member
Join Date: Sep 2013
Posts: 10

Originally Posted by ChipB 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 FfkT=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 FfkTumg, 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?

 
Sep 11th 2013, 06:29 AM

#9  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,347

Originally Posted by sadistprincess 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.
Last edited by ChipB; Sep 11th 2013 at 08:52 AM.

 
Sep 11th 2013, 11:11 AM

#10  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,347

Are you sure you have the complete problem? Is it from a text book? Perhaps they want the value of F+fk?

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