# How to calculate the angle using work, displacement and 2 forces?

• fixedglare
In summary, in a problem where a boat is pulled by a cord with an angle, the work done by the weight of the boat is 0.
fixedglare

## Homework Statement

Given:
Work = 12000 J
Weight: 800 N
displacement: 200 m
Force: 120 N
angle = ?

## Homework Equations

W = Fd

W = Fd cos θ

cos -1 (W/Fd) = θ

## The Attempt at a Solution

First I added the 2 forces together.

800 + 120 = 920 N

calculated work; 200 m * 920 N = 184000 J

then used the second formula;

cos -1 (12000 J /184000 J) = 86.3°

but the answer in my book says the answer is supposed to be 60°...

You're going to have to clarify the situation. What is going on in the problem? A block sliding, it looks like? Also, what do mean by the angle? The angle between what?

jackarms said:
You're going to have to clarify the situation. What is going on in the problem? A block sliding, it looks like? Also, what do mean by the angle? The angle between what?

The question says;

a person uses a cord to pull a boat with a mass of 1000 kg by 50.0 m along the board walk. The cord makes a 45 degrees angle with the horizontal force. If a force of 40 N is applied to the boat, how much work is realized?

I think you have the wrong problem -- in that one you're given the angle, but in this one you're trying to find it.

jackarms said:
I think you have the wrong problem -- in that one you're given the angle, but in this one you're trying to find it.

You're right, I got them mixed up.

The real one says,

12000 J of work is required to pull a wagon that weighs 800 N, in a distance of 200 m. If the applied force over the cord is 120 N, what is the angle?

Sorry for that.

No problem.

You have the same problem as before -- adding the forces to find the work, when they need to be done separately. So first find $W_{weight}$ and then $W_{app}$.

@Jacko: Please stop trying to lift boats and wagons. They stay level. No Wweight.

The point was to figure out that the work done by weight is zero.

Fixo: You're doing the right thing. W is work in your second relevant formula, not Weight. Don't let the weight or Jacko distract you. What is W/Fd ? right, no need for a calculator!

The first formula doesn't apply: F isn't in the direction of d. That's where the cos comes in: to calculate the component of F that is in the direction of d.

jackarms said:
The point was to figure out that the work done by weight is zero.

I would have never figured that out. But with this new found information I will try to do it again.

## 1. How do I calculate the angle using work, displacement and 2 forces?

To calculate the angle using work, displacement and 2 forces, you can use the formula: θ = cos^-1((F₁d₁ + F₂d₂)/W). Here, F₁ and F₂ are the two forces applied, d₁ and d₂ are the displacements caused by each force, and W is the work done.

## 2. What is the unit of measurement for the angle calculated using work, displacement and 2 forces?

The angle calculated using work, displacement and 2 forces is measured in radians (rad) or degrees (°), depending on the units used for the other variables in the formula.

## 3. Can the angle calculated using work, displacement and 2 forces be negative?

Yes, the angle can be negative depending on the direction of the forces and the displacement. A negative angle indicates that the forces are working against each other.

## 4. Is it necessary to know the magnitude of the forces in order to calculate the angle?

Yes, the magnitude of the forces is necessary to accurately calculate the angle using work, displacement and 2 forces. The formula requires the values of the forces to be included in the calculation.

## 5. Can the angle calculated using work, displacement and 2 forces be used to determine the direction of the forces?

No, the angle calculated using work, displacement and 2 forces only provides information about the relationship between the forces and the displacement. To determine the direction of the forces, additional information such as the coordinates of the forces may be needed.

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