# Free Body Diagrams on friction less surface

1. Aug 29, 2010

### dch1runs

1. A frictionless surface is inclined at an angle of 34.4° to the horizontal. A 270-g block on the ramp is attached to a 75.0-g block using a pulley, as shown in the figure below.
(a) Draw two free-body diagrams, one for the 270-g block and the other for the 75.0-g block. (b) Find the tension in the string and the magnitude of the acceleration of the 270-g block. (c) The 270-g block is released from rest. How long does it take for it to slide a distance of 0.90 m along the surface? (d) Will it slide up the incline, or down the incline?

F=m/a

Ok, so I'm having a lot of problem finding part (a). I'm pretty sure once I do, that b-d won't be very difficult at all. I rewrote the F=ma equation for both objects based on my free body diagrams. For the 75g object I had: a=g-T/m or a=9.81 m/(s*s) - T/.075g. For the 270g object i found that a=T/m-g*cos34.4. However when I try to solve as a system of equations my answers continue to be incorrect. Maybe the accelerations for the two objects are different? If so, then I have no idea where to go from there.

2. Aug 29, 2010

### Staff: Mentor

What's the component of the 270-g block's weight parallel to the incline?

3. Aug 29, 2010

### dch1runs

Its not given, but I think its 9.81cos34.4?

4. Aug 29, 2010

### Staff: Mentor

No. First off, the weight equals mg. mg*cosθ would be the component perpendicular to the incline.

5. Aug 29, 2010

### dch1runs

Does that mean that it would be mg/cos34.4?

6. Aug 29, 2010

### Staff: Mentor

No. Learn how to find the parallel and perpendicular components here: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]

Last edited by a moderator: May 4, 2017
7. Aug 29, 2010

### dch1runs

Thank you for that. So now I can find the perpendicular and parallel components. The parallel component would be m*g*sin(theta). Which in this example is .27kg*9.81m/(s*s)*sin34.4 I think. But its still not the magnitude of the acceleration, right? I know the site said to divide by the mass again, leaving the equation as g*sin(theta). But when I solve it, the application still tells me that I have the wrong answer. What else am I missing? Again thanks for your help.

Last edited: Aug 29, 2010
8. Aug 29, 2010

### Staff: Mentor

Good.
The acceleration would be g*sin(theta) if there were no other forces acting on the mass, but that's not the case here.
Identify the forces acting on each mass and set up your force equations:
ΣF = ma (for mass #1)
ΣF = ma (for mass #2)

Then combine the two equations to solve for the tension and the acceleration.

9. Aug 29, 2010

### dch1runs

So for the 75g object the forces acting on it are gravity and the rope.
F=m*a=m*g-T
The 270g object also has gravity (though only the parallel component) and the rope.
f=m*a=T-g*sin(theta)

But I still can't find the answer when putting them together and solving for tension and acceleration, are the accelerations different? If so then I only have two equations with three variables. I feel like I'm making this a lot more difficult than it really is.

10. Aug 29, 2010

### Staff: Mentor

Good. To make sure you keep the masses separate, write it as:
m1*a = m1*g - T
You left out the mass when you found the parallel component of the weight. Write it as:
m2*a = T - m2*g*sin(theta)

Now try combining them. (Hint: Just add them.)

11. Aug 29, 2010

### dch1runs

Alright, so I finally got it! Thank you, although I do have one general question with this type of problem. How is it that both objects have the same acceleration, is it because they are connected by the rope and pulley?

12. Aug 30, 2010

### Staff: Mentor

Exactly right. The connecting rope forces them to have the same acceleration. Cut the rope and they will move independently.

Note that in writing your equations you implicitly made use of that fact.