How can I solve a three-box system force problem?

[Ilyo]

Homework Statement

three boxes are slid across a frictionless surface, 100N is applied to m1(mass1) which is pushed against m2 and m3 each box weighs 10 kg
a draw a free body diagram for each box
b find the acceleration of the system
c the net force on each block
d the force of contact that each block exerts on its neighbor

Homework Equations

[sigma]F=ma

The Attempt at a Solution

Well I have drawn my free body diagrams


then I set up my equation for part b
though I am not sure where to go from there I was hoping somone could show me how to do the rest of the problem with different values than the ones I was given, and explain how they did it so I can get a better grasp on what to do here..


{now since it is a frictionless surface and a normal force pushes back as much as is applied to it would the third objects normal force be 100 Newtons? If so then it would be really simple and turn out to bed
NF=ma
100=10a} <---I understand this is wrong...

I feel way off track?

Thanks for your insights

 

[ehild]

Make a picture with a drawing program and upload as attachment.

Draw the normal forces in your figure. The boxes are in contact, so they move together with the same acceleration. Apply sum(F) =ma for each block.


ehild

 

[Ilyo]

fixed...

 

[ehild]

The first block is pushed with Fa. Fa can not act between two blocks.

Two boxes act to each other with forces of equal magnitude and opposite directions.

The boxes move on a horizontal surface, their weight, mg, cancels with the normal force of the surface. You need to work only with the horizontal forces .
There is no normal force at the free end of the third box.ehild

 

[Ilyo]

if there is no normal force at the end of the last box then what is pushing it forward?

 

[ehild]

if there is no normal force at the end of the last box then what is pushing it forward?

The other box at the other side.

ehild

 

[Ilyo]

right right.. I think I got it now thanks ^^