Is this Free-Body Diagram Correct?

[DevonR]

Homework Statement

I have to draw a free body diagram representing the motion of an object @ a given point in time. This diagram is for section 1.

It's of a 1.25 kg cart with a 0.6 kg can attached to it with a pulley-system. When the can drops, the cart begins to move. In section one, when the can drops, the cart begins to accelerate forward.

Acceleration = 0.196 m/s(squared)
Mass of cart = 1.25 kg
Mass of can = 0.6 kg

Homework Equations

F = ma

The Attempt at a Solution

The diagram is uploaded here. I just want to know if this seems to be correct?

EDIT:

Heres the assmuptions I made

I calculated net force w/ acceleration provided = 0.196 m/s
and with the total mass (1.25 kg + 0.6 kg) = 1.9 kg
Net force = 0.37 N

I did normal force & force of gravity on the cart, which was 12.3 N for both.
The falling can was what caused the motion of the cart, so I calculated Fg = 5.9 N for the 0.6 kg can. No normal force obviously because it wasn't sitting on a surface.

But the can was attached to a string, which slowed down the falling can's motion. So I had to calculate F-tension. I used this formula to calculate it;

Force of Tension = 2m₁m₂/m₁m₂ x Gravity

I got 4N (3.97N) as the tension force.

So I assumed that the applied force would be Force of Gravity on the can MINUS the Force of Tension of the string holding the can. The applied force was 1.9 N

Then I calculated force of friction.
I assumed it to be the applied force minus net force
Which was 1.9 N - 0.37 N

For friction I got 1.5 N.

Does all of this seem correct? [/B]

 

[haruspex]

I would have thought the set up would be so that the falling can moves the cart horizontally. As you've drawn it, the force on the cart is vertical.

 

[DevonR]

Thanks for the reply! I added a written description of my 'attempt @ a solution' to the original post.
I see what you mean, so I should draw Fa = 1.9 horizontally.

Aside from that, is everything else okay?

 

[haruspex]

I got 4N (3.97N) as the tension force.

How did you get that? Without the tension, the acceleration would be g. So what tension is required to reduce it to the given acceleration?

 

[DevonR]

How did you get that? Without the tension, the acceleration would be g. So what tension is required to reduce it to the given acceleration?

I used this formula Force of Tension = 2m₁m₂/m₁m₂ x Gravity
to get the force of tension

 

[haruspex]

I used this formula Force of Tension = 2m₁m₂/m₁m₂ x Gravity
to get the force of tension

I've no idea where you get that equation from. All the m's seem to cancel.

 

[DevonR]

I've no idea where you get that equation from. All the m's seem to cancel.

I got the formula from here

http://www.citycollegiate.com/tension1.htm

 

[haruspex]

I got the formula from here

http://www.citycollegiate.com/tension1.htm

Two problems:
- you omitted the plus signs, minus signs and parentheses, making it impossible to decipher.
- it doesn't apply in your set up; compare the diagrams
It's quite easy from first principles without searching for answers on the net. Just consider the FBD of the hanging mass.