# Finding Acceleration in two Force problems

1. Oct 13, 2008

### jllorens

1. The problem statement, all variables and given/known data for problem 1
A 2.50 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object, as shown in the figure. Find the magnitude of the acceleration of the two objects and the tension in the string.

2. Relevant equations for problem 1
F=ma

3. The attempt at a solution for problem 1
I have absolutely no idea how to go about solving this. Originally, I was thinking that the acceleration would simply be gravity, since the table is frictionless and there is no friction acting on the block sitting on the table to inhibit the downward motion of the hanging block. Obviously, that is wrong.

1. The problem statement, all variables and given/known data for problem 2
A 2.60 kg object is moving in a plane, with its x and y coordinates given by x = 4t2 - 1 and y = 4t3 + 4, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.45 s.

2. Relevant equations for problem 2
F=ma

3. The attempt at a solution for problem 2
I originally took the first derivative of each component equation, inserted the given value for t, and then did the following to obtain a resulting number from both components: a = sqrt(x^2+y^2). Obviously I am quite lost. Am I on the right track by using derivatives? Should I be putting the y equation over the x equation (rise over run, slope) and taking the first derivative of that?

2. Oct 13, 2008

### Epsillon

So Fnetsystem=(m1+m2) $$\times$$ $$\vec{}a$$

2 for bigger object 1 for smaller

Fnetsysten= Fg2+Ft1-Ft2 Ft1=Ft2
Fnetsysten= Fg2
Fnetsysten= m2g

so m2g==(m1+m2) $$\times$$ $$\vec{}a$$
(9)(9.8)==(9+2.5) $$\times$$ $$\vec{}a$$
$$\vec{}a$$ = 7.67

Now that you have a u can put it back in the formulas to get the tension forces.
I am not positive about my answer but this is how I would do it.

3. Oct 13, 2008

### jllorens

I am having trouble understanding what you did here.

4. Oct 13, 2008

### Epsillon

Well in there I just wanted to show how the Tension forces cancell out.

There are two tension forces: one going against the direction of the acceleration (acting upwards on the big mass) and one going in the direction of acceleration (pulling the object on the table towards the hanging mass) the two forces are the same since the same rope is used; therefore, they cross out leaving us with the gravity force.

5. Oct 13, 2008

### jllorens

The correct answer for the tension in the string however is 19.175. If the opposing tensions canceled, would that not mean that the tension in the string is 0? I got 19.175 when I multiplied M1 times a, the force of block 1 (the block on the table). I am not sure, however, why that is.

6. Oct 13, 2008

### Epsillon

Yes they cross out for the Fnet but that doesent mean they dont exist.

To find Ft1 I would focus on one of the specific objects.

So for the one on the table

Fnet= Ft

ma= Ft
2.5 x 7.67 = 19.175

You can also look at the second object (hanging mass)

Fnet= Fg-Ft
ma=mg- Ft
(9 x 7.67) = (9 x 9.8 ) -ft

Once again we get 19.175

proving that the tensions forces are the same.