# Understanding Tension on a Pulley

• TiredCollegeStudent
In summary, the problem requires finding the reaction forces at point E, specifically the Fx, Fy, and Moment. The given data includes 20kN concentrated forces at load AC, with a distance of 1.8m between them, and a tension of 150kN in the cable. The solution provided does not include the force components at the BD part of the rope, as it is considered part of the internal system and only external forces such as gravity and ground at points E and F are relevant. This is because internal forces always consist of equal and opposite pairs, resulting in no net force or moment on the system. The subsystem consisting of the cable alone can also be considered in equilibrium.
TiredCollegeStudent

## Homework Statement

Problem image: https://prnt.sc/gvhjso
In this case I have to find the reaction forces at the point E (the Fx, Fy, and the Moment at point E) by using the given data. The 20kN forces at the load AC are concentrated, and are 1.8m far from each other. The tension of the cable is 150kN

## Homework Equations

Fx = Fcos( θ w/ respect to x-axis) or;
= Fsin( θ w/ respect to y-axis)
Fy = Fsin( θ w/ respect to x-axis) or;
= Fcos( θ w/ respect to y-axis)
Moment = Fd (where d is the perpendicular distance between the point of moment and the force)

## The Attempt at a Solution

This problem has the solution already given, which is this:
https://prnt.sc/gv93sy

Now my problem is I do not understand why the solution did not include the force components at the BD part of the rope, only factoring the force components of the DF part of the rope/string. Why is it not included in solving the missing reaction forces at point E?

Last edited by a moderator:
Arnav jamale
According to the FBD in the solution, the system consists of the vertical and horizontal booms plus the rope. The BD part of the rope is part of this system and the system cannot exert a force on itself. Only the parts of the world that are external to the system can exert a force or torque on it. These are Earth (gravity acts "at a distance") and the ground at points E and F.

TiredCollegeStudent
kuruman said:
According to the FBD in the solution, the system consists of the vertical and horizontal booms plus the rope. The BD part of the rope is part of this system and the system cannot exert a force on itself. Only the parts of the world that are external to the system can exert a force or torque on it. These are Earth (gravity acts "at a distance") and the ground at points E and F.

So you mean to say that the BD part of the rope still exerts a force on its own but that force is not relevant to the whole system? I find it hard to accept because should the force exerted by the BD part of the rope become too low, won't the boom AC fall down, assuming its weight is too heavy for the support at point C to handle alone?

TiredCollegeStudent said:
So you mean to say that the BD part of the rope still exerts a force on its own but that force is not relevant to the whole system? I find it hard to accept because should the force exerted by the BD part of the rope become too low, won't the boom AC fall down, assuming its weight is too heavy for the support at point C to handle alone?
As kuruman wrote, the system being considered is the frame plus the cable. Internal forces are irrelevant because they always consist of equal and opposite pairs acting on that system.
E.g. we could choose to consider the cable BD as a separate system. The tension in BD results in a force on the rest of the frame+cable system at B and an equal and opposite force on that subsystem at D, and these are in the same line, so no net force and no net moment.

TiredCollegeStudent
So it's like this: There is a tension from point B and there is also a tension from point D and they both cancel each other out? So this subsystem is also considered an equilibrium system on its own? Is this assumption right?

TiredCollegeStudent said:
There is a tension from point B and there is also a tension from point D and they both cancel each other out?
Yes.
TiredCollegeStudent said:
So this subsystem is also considered an equilibrium system on its own?
It can be. You are free to subdivide or combine into whatever systems you wish. If all is static then each must be in equilibrium in its own right.

TiredCollegeStudent
Alright, thanks for answering, kuruman and haruspex :D

## What is tension on a pulley?

Tension on a pulley refers to the force that is applied to the rope or belt that runs over the pulley. It is the force that is transmitted through the rope or belt, and can be measured in Newtons.

## How does tension affect the motion of a pulley?

Tension plays a crucial role in the motion of a pulley. If the tension is equal on both sides of the pulley, the pulley will remain stationary. However, if the tension on one side is greater than the other, the pulley will move in the direction of the greater tension.

## What factors affect tension on a pulley?

The tension on a pulley is affected by several factors including the weight of the object being lifted, the angle of the rope or belt, and the friction of the pulley. The greater the weight or angle, and the more friction present, the greater the tension will be.

## How is tension calculated on a pulley system?

To calculate tension on a pulley system, you can use the formula T = (m x g) + (F x sinθ), where T is the tension, m is the mass of the object, g is the acceleration due to gravity, F is the force applied to the pulley, and θ is the angle of the rope or belt.

## What are some real-world applications of understanding tension on a pulley?

Understanding tension on a pulley is important in various fields such as engineering, physics, and mechanics. It is used in lifting and transportation systems, as well as in simple machines like elevators and cranes. It is also used in experimental setups for measuring forces and in everyday tasks like lifting heavy objects using a pulley system.

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