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

[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°...

help me please.

 

[jackarms]

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?

 

[fixedglare]

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?

 

[jackarms]

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.

 

[fixedglare]

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.

 

[jackarms]

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 $W_{app}$.

 

[BvU]

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

 

[jackarms]

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

 

[BvU]

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.

 

[fixedglare]

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.