# Calculating Tension with Kinetic Friction in a Pulley

• limpexowns
In summary, the block has static friction and gravity and when it is pushed from the back it has kinetic friction.
limpexowns

## Homework Statement

(1) Firstly, the block as it rest at incline of 10 degrees and mass is added on the other side of pulley to overcome static friction (0.0063kg). The block weighs 0.8kg. Calculate the tension.

(2) The block is at same angle and mass but this time the block was lightly pushed from back to make it move. As the block began to move, mass was added on the other side of the pulley to just keep moving it forward. The mass was (0.00480kg). Calculate Tension.

## Homework Equations

T=mg (to calculate tension for static friction)
T=mg+Fs
Fs=T-Fgsin@
Fk=T-Fgsin@

## The Attempt at a Solution

For (1), I used T=mg to find tension which comes out to be 0.618N. I used that value to find Static Friction and I got 0.745N

For (2), I tried using T=mg to find Tension and I got 0.471N but when I used that value to find Kinetic friction, I'm getting 0.892N. This value didn't make sense because the Kinetic friction is always less than static friction but here it was higher.

So, I tried using T=mg+Fs for tension in Kinetic friction. I assumed that Fs was force applied because the cart was slightly pushed just o make it move. By using that, I got T=1.22N. I used that value to find Kinetic friction and I'm getting 0.143N. This value makes sense to me because the kinetic friction is lower than static.

The questions are very similar to this lab: http://ems.calumet.purdue.edu/chemphys/ncrelich/PortableDocuments/PHYS152/Summer2009/LabInclinedPlane.pdf

So, can you please let me know which method is right in order to calculate Tension for Kinetic part. Thanks a lot.

Last edited by a moderator:
(1) Firstly, the block as it rest at incline of 10 degrees and mass is added on the other side of pulley to overcome static friction (0.0063kg). The block weighs 0.8kg. Calculate the tension.

Then the tension is g*0.0063kg+g*0.8kg*sin10°
Do you agree ?

Since the mass on the pulley is the one that made the block start moving. I think that the tension produced by that mass includes Force of gravity in the X and static friction of the block. That's why the equation I used is T=mg.

## What is tension?

Tension is the pulling force exerted by a string, rope, or cable on an object. It is a measure of the amount of force being applied to keep an object in place or to move it.

## How is tension calculated?

Tension can be calculated by using the equation T = mg, where T is the tension force, m is the mass of the object, and g is the acceleration due to gravity (9.8 m/s^2). This equation assumes that the object is at rest or moving at a constant velocity.

## What is kinetic friction?

Kinetic friction is the force that resists the motion of an object when it is in contact with a surface. It is caused by the microscopic irregularities on the surface of the object and the surface it is in contact with. Kinetic friction is always present when there is relative motion between two surfaces.

## How does kinetic friction affect tension in a pulley system?

In a pulley system, the tension in the rope or cable is affected by the presence of kinetic friction. As the object moves, the friction force acts in the opposite direction of the motion, reducing the tension in the rope or cable. This reduction in tension is proportional to the coefficient of kinetic friction and the weight of the object.

## How do you calculate tension with kinetic friction in a pulley system?

To calculate tension in a pulley system with kinetic friction, you can use the equation T = mg + μk * m * g, where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, and μk is the coefficient of kinetic friction between the object and the surface it is in contact with. This equation takes into account the reduction in tension caused by the presence of kinetic friction.

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