Unbalanced Forces on Ramp with Tension

[RivalDestiny]

Homework Statement

The suspended 2.3 kg mass on the right is moving up, the 2.4 kg mass slides down the ramp, and the suspended 7.2 kg mass on the left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9.8 m/s2. The pulleys are massless and frictionless.

What is the tension in the cord connected to the 7.2 kg block? Answer in units of N.

Here is a photo:

Homework Equations

F=ma
Fg=mg
$$\mu$$=Ff/FN

The Attempt at a Solution

I split it up into gravity parallel and gravity perpendicular and did the following to get those values:

sin23 * 23.52 = 9.19 N (parallel)
cos23 * 23.52 = 21.65 N (perpendicular)

Since the perpendicular force is equal to the normal force, the value of the normal force is also 21.65 N. Then I plugged the following to find the friction force:

$$\mu$$=Ff/FN

0.12 = Ff/21.65
Ff = 2.598 N

Then, I found the gravitational forces on the blocks hanging on the two sides.

7.2 kg * 9.8 m/s^2 = 70.56 N
2.3 kg * 9.8 m/s^2 = 22.54 N

This is the part I'm currently stuck on. Do I add up all the forces on the 2.4 kg block? Or do I have to use a summation to find the tension in the 7.2kg string?

 

[Jolb]

The easiest way to do this problem is to find the equation of motion for the mass on the ramp. Call that mass m'.

Do vector addition for the forces on m'. The force from the ropes are in opposite directions, so they can be added easily. Call the heavier one m1 and the lighter m2. You can then find the acceleration of mass along the ramp.
$$a_{m'}=\frac{1}{m'}((m1-m2)g\sin \theta -\mu m'g\cos\theta)$$

Where theta is 23 degrees.
The term with the mu is the frictional force. Remember that for calculating the total tension in the left rope.

Now you have the acceleration of m', which is equal in magnitude to the accelerations of all the boxes. Remember that the downward force on the hanging boxes is not mg because they are accellerating: Fdown = m(g-a)

Note:I just edited that equation. By accident I left the right hand as the force--now it's devided my m' to be the correct acceleration.

 

[kerimek]

I would like to clarify a few things before providing a response. Firstly, it is important to specify the direction of motion for each of the masses. From the given information and the photo, it seems like the 2.3 kg mass is moving up, the 2.4 kg mass is moving down the ramp, and the 7.2 kg mass is moving down. Secondly, it would be helpful to know the angle of the ramp to accurately calculate the forces involved.

Assuming that the 2.3 kg mass is moving up at a constant velocity, it means that the net force acting on it is zero. This also means that the tension in the cord connected to the 2.3 kg mass is equal to the weight of the mass, which is 2.3 kg * 9.8 m/s^2 = 22.54 N.

For the 2.4 kg mass, since it is moving down the ramp, there must be a net force acting on it in the downward direction. This force can be broken down into two components - the force of gravity acting on the mass (2.4 kg * 9.8 m/s^2 = 23.52 N) and the friction force (Ff). Using the equation F=ma, we can calculate the acceleration of the mass as 23.52 N/2.4 kg = 9.8 m/s^2. Since the mass is moving down the ramp, the direction of acceleration is also down the ramp. Therefore, we can use the equation F=ma again to calculate the net force acting on the mass in the downward direction (F=2.4 kg * 9.8 m/s^2 = 23.52 N). This means that the friction force must be equal to the net force acting on the mass, which is 23.52 N - 23.52 N = 0 N. This suggests that the friction force is not enough to prevent the mass from sliding down the ramp.

For the 7.2 kg mass, since it is also moving down the ramp, there must be a net force acting on it in the downward direction. This force can be broken down into two components - the force of gravity acting on the mass (7.2 kg * 9.8 m/s^2 = 70.56 N) and the tension in the cord connected to the mass (T