Friction & Forces - Calculate Force on Wheaties Box

[thegoosegirl42]

Homework Statement

A box of Cheerios and a box of Wheaties are accelerated across a horizontal surface by a horizontal force F applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2N, and the magnitude of the frictional force on the Wheaties box is 4N. If the magnitude of F is 12N, what is the magnitude of the force on the Wheaties box from the Cheerios box.

Homework Equations

Friction=uN
F=ma[/B]

The Attempt at a Solution

a=F/m
12-2-4=6
a=6/(C+W) ----- I am using C and W to represent the mass of the different boxes.
Cgu=2
Wgu=4
However I don't know where to use substitution on this.

 

[Chestermiller]

If the coefficient of kinetic friction is the same for both boxes, how does the mass of the cheerios box compare with the mass of the wheaties box? Once you know this, draw a free body diagram of the wheaties box, and write down the force balance equation for the wheaties box.

Chet

 

[thegoosegirl42]

If the coefficient of kinetic friction is the same for both boxes, how does the mass of the cheerios box compare with the mass of the wheaties box? Once you know this, draw a free body diagram of the wheaties box, and write down the force balance equation for the wheaties box.

Chet

I didn't think that the frictional forces were equal but if they are the mass of the bigger one is twice that of the smaller one.

 

[Chestermiller]

I didn't think that the frictional forces were equal but if they are the mass of the bigger one is twice that of the smaller one.

The equations you wrote in post #1 imply that you do think the coefficients of friction are equal.

Let's see the force balance from your free body diagram of the wheaties box.

Chet

 

[CWatters]

I didn't think that the frictional forces were equal but if they are the mass of the bigger one is twice that of the smaller one.

Chet said "coefficient of kinetic friction" not "frictional forces". There is a difference.

 

[thegoosegirl42]

Chet said "coefficient of kinetic friction" not "frictional forces". There is a difference.

Yeah, I meant coefficient of friction but I didn't realize until I replied and by then it was too late. :-(

 

[thegoosegirl42]

The equations you wrote in post #1 imply that you do think the coefficients of friction are equal.

Let's see the force balance from your free body diagram of the wheaties box.

Chet

Thank you so much. I guess I was just reading too far into the problem.