Finding force F when R equals 234 lb

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funlord
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Homework Statement



Textbook figure P2.20: a crate on a ground line with three blue arrows labeled F pulling from its right side; one points up-right at 25 degrees, one points right, and one points down-right at 45 degrees.


Homework Equations


R=Rxi+Ryj

The Attempt at a Solution


I think it has insufficient given but the answer at the back is 234lb.
I assumed that when i put a line through the angle bisector, the line would be perpendicular to the crate

PS: IF the image doesn't load, i put an attachment of the image
Mod note: I have edited your post so that the image now shows.
 
Last edited by a moderator:
Edit - I am wrong, sorry.

I think the problem has sufficient information. When you find the equivalent force, you will find an angle and magnitude.
 
funlord said:

Homework Statement



https://lh3.googleusercontent.com/B-WBHAU9CTvs83wBK4Zcz8NZOtxHirzD8U8hWFAPXw0I2YRYlwiPacM6mG_oTzfci79cjqNj=w1366-h768-rw-no

Homework Equations


R=Rxi+Ryj[/B]

The Attempt at a Solution


I think it has insufficient given but the answer at the back is 234lb.
I assumed that when i put a line through the angle bisector, the line would be perpendicular to the crate

PS: IF the image doesn't load, i put an attachment of the image

First, create a diagram, if you haven't already done so and use the method you think will lead you to the resultant force. What method would you use?
 
I say that the sum of the x component has to be equal to the force required, and that of the y componet has to be 0 so here is a system of equations that satisfy this requirements:
(I called X the force of 600lb)
 
Attachments
  • Handwritten system of two equations: F cos(25°+θ) + F cos θ + F cos(45°−θ) = X and F sin(25°+θ) + F sin θ − F sin(45°−θ) = 0, above a handwritten diagram of three forces F at angles 25°, θ and 45° from the x-axis.
    20160706_170213.jpg
    23.8 KB · Views: 610
Cozma - you need to retract your like. ;-) I was incorrect.

Break the problem into multiple steps - the first step is to combine the forces as drawn to a single force, just a QuantumQuest suggested..
 
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It doesn't say that the crate will be pulled horizontally. Therefore we can't assume that y- component is 0.
 
Cozma Alex said:
I say that the sum of the x component has to be equal to the force required, and that of the y componet has to be 0 so here is a system of equations that satisfy this requirements:
(I called X the force of 600lb)

Then what will be next?
There are three unknowns, therefore it requires three equations
 
funlord said:
Then what will be next?
There are three unknowns, therefore it requires three equations
No X is 600 lb, I just called it X
 
QuantumQuest said:
First, create a diagram, if you haven't already done so and use the method you think will lead you to the resultant force. What method would you use?

I tried the best of what I can do, but it doesn't lead to a possible solution
 
Cozma Alex said:
No X is 600 lb, I just called it X

ahh ok, I will try to solve it, if it leads that F is 234lb. Then that's the correct solution
 
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funlord said:
Then what will be next?
There are three unknowns, therefore it requires three equations

You are getting ahead of yourself. First - pretend the problem statement is -

"Write an equation for a single equivalent force f in terms of F and the angles given. What is the angle of the resultant force f?"

You have enough information in the problem to solve the above. Once you do that, figuring out how to set F to make the magnitude of f any value you want should be clear.
 
funlord said:
I tried the best of what I can do, but it doesn't lead to a possible solution

What was that best? Think on your own what you should use to solve the problem and why.

I kindly request to respect PF rules and not give full solutions. There's absolutely no point to do that. No one learns this way.