# Unbalanced Forces on Ramp with Tension

• RivalDestiny
In summary, the conversation discusses the setup of a physics problem involving three masses connected by ropes and a ramp, with friction present. The goal is to find the tension in the rope connected to the 7.2 kg block. The solution involves finding the equation of motion for the mass on the ramp and then using vector addition to find the forces on that mass. From there, the acceleration of the mass can be calculated, which is equal to the acceleration of all the blocks. Finally, the tension in the rope can be found by considering the downward force on the hanging blocks.
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?

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.

Last edited:

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

## 1. What is an unbalanced force on a ramp with tension?

An unbalanced force on a ramp with tension refers to a situation where the forces acting on an object on a ramp are not equal, causing the object to accelerate in a certain direction.

## 2. How does tension affect the motion of an object on a ramp?

Tension is a force that is transmitted through a rope, cable, or string that pulls on an object. In the case of a ramp, tension can either assist or resist the motion of an object, depending on the direction of the force.

## 3. What is the relationship between the angle of the ramp and the tension force?

The angle of the ramp affects the tension force in that a steeper angle will result in a greater tension force. This is because a steeper ramp requires more force to overcome the increased gravitational force pulling the object down the ramp.

## 4. How does mass affect the tension force on a ramp?

The mass of an object does not directly affect the tension force on a ramp. However, mass does affect the overall motion of the object, which in turn can affect the tension force. A heavier object may require a greater tension force to move it up a ramp compared to a lighter object.

## 5. Can an object on a ramp experience both tension and friction forces?

Yes, an object on a ramp can experience both tension and friction forces at the same time. Tension acts in the direction of the rope or cable, while friction acts in the opposite direction of motion. These forces can work together or against each other to affect the motion of the object on the ramp.

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