Finding Acceleration in two Force problems

[jllorens]

1. Homework Statement 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. Homework Equations for problem 1
F=ma3. 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. Homework Statement 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. Homework Equations for problem 2
F=ma3. 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?

 

[Epsillon]

1. Homework Statement 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. Homework 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.

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.

 

[jllorens]

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

I am having trouble understanding what you did here.

 

[Epsillon]

I am having trouble understanding what you did here.

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.

 

[jllorens]

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.

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.

 

[Epsillon]

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.

Yes they cross out for the Fnet but that doesent mean they don't 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.