W = F d cos θ Is Algebra Okay for Work?

[peacerosetx]

Greetings all

Homework Statement

Greetings all this is the basic type of problem that I am having. W = F d cos θ. This is not for homework this is for my preparation for an exam.

If you were pulling a box like in the example above, which moves 12.7 m when you pull
along the rope with a force of 76.0 N, determine how much work you did.

Homework Equations

it has to do with the Work:: W = F d cos θ

The Attempt at a Solution

I have the answer which is given as:

W = F d cos θ
= 76.0 N (12.7 m) cos30.0°
W = 836 J

The problem is that I do not know how to multiply cos30.0 to know how you get 836 J

I have tried to turn it into .5 so that it would be = 76.0 N(12.7m) (.5) but when I multiply it all together I get 482 and not 836 J

Can you please help me figure this out is it the way I am doing the algebra that is not correct or my conversion of the cos 30? Thank you

 

[cepheid]

Welcome to PF.

The cosine of 30 degrees is NOT 1/2. That's why you didn't get the right answer. Generally speaking, All you're expected to do is to use a calculator to figure out what cos(30^o) is.

 

[peacerosetx]

Welcome to PF.

The cosine of 30 degrees is NOT 1/2. That's why you didn't get the right answer. Generally speaking, All you're expected to do is to use a calculator to figure out what cos(30^o) is.

Yes but for this exam that I will take I am not allowed to use a calculator. And I really would like to know how to use cosine and multiply with it not only for this but for other things as well. Do you know how to get to get to836 J without a calculator, or is it too hard to compute without a calculator please? Thank you. The examples seem to feel that we can do it on a scratch piece of paper. Thank you for your response

 

[cepheid]

The sine and cosine of 30° (π/6 radians) and 60° (π/3 radians) are supposed to be easy to remember or derive because in this situation you have a special right triangle known as a 30°-60°-90° right triangle:

http://en.wikipedia.org/wiki/Special_right_triangles#30-60-90_triangle

By the way, your algebra and method is fine. You just had the wrong numbers.

 

[peacerosetx]

The sine and cosine of 30° (π/6 radians) and 60° (π/3 radians) are supposed to be easy to remember or derive because in this situation you have a special right triangle known as a 30°-60°-90° right triangle:

Thank you for the information and the welcome I am still confused as to how to solve the equation. I will work on it some more though. God bless

 

[Redbelly98]

Yes but for this exam that I will take I am not allowed to use a calculator.

Then how did you multiply (76.0 N) * (12.7 m)? I guess you could have done it by hand, but I don't think a physics exam that doesn't allow calculators would ask you to do that.