# Resolving forces: Mass on a string

• thebosonbreaker
In summary, the mass hangs from a string equilibrium. A force of 0.3N is applied, making an angle of 30 degrees with the horizontal. The tension in the string is found to be 0.05g.
thebosonbreaker

## Homework Statement

A mass of 50 grams hangs in equilibrium on a string. The mass is pulled aside and upwards by a force of 0.3N which makes an angle of 30° with the horizontal. Find the angle that the string makes with the vertical and the tension in the string.

## Homework Equations

Body is in equilibrium, so I will need to resolve forces (tension and weight).

## The Attempt at a Solution

For the first part of the question (find the angle that the string makes with the vertical):
I do not understand why this is not 60° - the string will move in the same direction as the 0.3N force applied to it, so if it makes an angle of 30° with the horizontal (the angle between 'horizontal'/'vertical' being a right angle), the angle made with the vertical must be: 180 - (90 + 30) = 60°. Could somebody please explain why this is not the case and how to find the angle?

For the second part of the question (finding the tension in the string):
Resolving horizontally I have that: Tcos30° = 0.3 [the horizontal component of tension equals the horizontal force applied]
Resolving vertically I have that: Tsin30° = mg = 0.05g

Could somebody clraify this problem? I'm having a bit of a hard time with it.

thebosonbreaker said:

## Homework Statement

A mass of 50 grams hangs in equilibrium on a string. The mass is pulled aside and upwards by a force of 0.3N which makes an angle of 30° with the horizontal. Find the angle that the string makes with the vertical and the tension in the string.

## Homework Equations

Body is in equilibrium, so I will need to resolve forces (tension and weight).

## The Attempt at a Solution

For the first part of the question (find the angle that the string makes with the vertical):
I do not understand why this is not 60° - the string will move in the same direction as the 0.3N force applied to it, so if it makes an angle of 30° with the horizontal (the angle between 'horizontal'/'vertical' being a right angle), the angle made with the vertical must be: 180 - (90 + 30) = 60°. Could somebody please explain why this is not the case and how to find the angle?

For the second part of the question (finding the tension in the string):
Resolving horizontally I have that: Tcos30° = 0.3 [the horizontal component of tension equals the horizontal force applied]
Resolving vertically I have that: Tsin30° = mg = 0.05g

Could somebody clraify this problem? I'm having a bit of a hard time with it.
In equilibrium, the string aligns in the direction of the resultant of gravity and the applied force.

If the object were heavy enough the applied force might only move it a very short distance. No where near 60 degrees.

The thread title mentions a spring. I don't see any mention of a spring in the problem itself.

Edit: Does the thread title need correction?

Have you drawn a free body disgram showing the forces acting on the mass, or do you feel like you have advanced beyond the point where you need to use free body diagrams?

## What is the concept of resolving forces?

The concept of resolving forces involves breaking down a single force vector into its component vectors along different axes. This is done in order to better understand how these forces act on an object.

## How do you resolve forces?

To resolve forces, you must first determine the direction and magnitude of the force vector. Then, using trigonometric functions, you can break down the force vector into its horizontal and vertical components.

## What is the importance of resolving forces?

Resolving forces allows us to better understand the overall effect of multiple forces acting on an object. It also enables us to accurately calculate the net force and acceleration of an object.

## What is the difference between a balanced and unbalanced force?

A balanced force is when the net force on an object is equal to zero, meaning there is no overall acceleration. An unbalanced force, on the other hand, results in a net force that is not equal to zero, causing the object to accelerate in the direction of the net force.

## How is resolving forces used in real-world applications?

Resolving forces is used in various fields such as engineering, physics, and mechanics. It is essential in understanding the forces acting on structures, vehicles, and other objects in order to ensure their stability and safety.

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