# Homework Help: Determining unknown Force

1. Feb 16, 2015

### Jose Miranda

1. The problem statement, all variables and given/known data
A 3.7 kg mass has a constant acceleration from an initial velocity of 2.33 ihat m/s to a final velocity of (18.9 ihat - 10.7 jhat) m/s in 2.71 s. Two forces act on the mass: gravity and an unknown applied force. Determine the acceleration and unknown force.

2. Relevant equations

3. The attempt at a solution
For the acceleration i managed to get 16.25 m/s, and for the force of gravity it is -36.26 m/s.

2. Feb 16, 2015

### BvU

Hello Jose, welcome to PF

Can you show your attempt ? The numerical value of the answer isn't as important as how you found it.
And what equation would you like to use to determine the applied force ? I assume that's where you run into problems ? Where precisely ?

3. Feb 16, 2015

### Jose Miranda

I believe the a I got the first time is wrong.
a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s

The problem is on on incline starting at 0,0 and having an angle below the x-line.

So the force on the x axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N

22.6 +14.6 = 37.2 N

37.2N - 36.26 N = 0.94 N
0.94 N = unknown force

4. Feb 16, 2015

### lightgrav

gravity around here, all by itself, causes acceleration 9.8 m/s2 ... that's 32 ft/s2 , if anybody still uses feet.
So the Force by gravity is 36.26 Newtons ... watch your units.

5. Feb 16, 2015

### jbriggs444

I agree with the corrected numbers. Thank you for showing your work.

22.6 N in one direction plus 14.6 N in the direction at right angles equals what?

Same problem. 37.2 N in what direction? Minus 36.26 N in what direction?

6. Feb 16, 2015

### Jose Miranda

What stumps me is the how can i do this if there is no angle.

7. Feb 16, 2015

### jbriggs444

You have referred to an "incline", an "x-axis" and a point at "0,0". Possibly there is a diagram that has not been shared? A reasonable assumption would be that gravity acts in the -y direction. You do not need an angle if you consider gravity and the y component of the unknown force together and the x component of the unknown force separately.

8. Feb 16, 2015

### haruspex

The problem statement is still unclear to me.
If it is 'on an incline', there would be a normal force from the surface. Is that the unknown force, or considered a consequence of the gravity?
Please state the entire question word for word.

9. Feb 16, 2015

### Jose Miranda

The problem doesn't mention an incline i just assumed it was on an incline, because of its final velocity.

10. Feb 16, 2015

### BvU

I subscribe to the a = (6.1ihat - 3.95 jhat) m/s, and also to force on the x axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N
but from there you start losing me:

If you go one step to your right ($\hat\imath$) and one step forward ($\hat\jmath$), the result is that your displacement is √2 steps at 45 degrees.
It's called vector addition. The two forces have to be added as vectors -- if you want to add them at all. But in this exercise you don't want that. They give velocities as $\ ( a\; \hat\imath + b\; \hat\jmath\ )\$, so an answer in the same terms should be OK.​

The gravity force that works in the $\hat\jmath$ direction is $\ m\;\vec g\ = 3.7 * -9.81\; \hat\jmath\$. If the net force is $\ -14.6\; \hat\jmath\$ N (I hope you didn't forget the minus sign ?) then the applied force in the y-direction must be quite positive.

This whole escapade can take place on an inclined plane or in thin air -- if the exercise tekst says "Two forces act on the mass: gravity and an unknown applied force" then that is enough to calculate that one unknown force. Which is all the exercise wants you to determine.

11. Feb 16, 2015

### Jose Miranda

ok, so an answer for this can be given as vectors? (22.60ihat N -14.6 jhat N) ?

12. Feb 16, 2015

### BvU

What you describe here is the net force. That isn't the answer for this exercise. I think the exercise asks for the second force.