Statics problem on the resolution of forces

[CIA16]

Homework Statement

I've tried solving the question and was stuck after creating an equation. I think my approach was wrong.

Relevant Equations

This is the equation I got when I resolved the forces and introduced the inequality symbol.

Here is my interpretation of the question. Three forces are applied to a bracket such that a 500N force acts at 30 degrees to the horizontal on the negative x-axis, while a 150N force acts at θ to the negative x-axis and another 150N force acts at 50+θ to the negative x-axis. The directions of the two 150N forces may vary, but the angle between the forces is always 50. Determine the range of theta values for which the magnitude of the resultant force acting at the bracket is less than 600N.
I already resolved the forces and introduced the inequality:
$$\sqrt{(-500 \cos(30) - 150 \cos(x) - 150 \cos(50 + x))^2 + (500 \sin(30) - 150 \sin(x) - 150 \sin(50 + x))^2}<600$$

 

[erobz]

Homework Statement: I've tried solving the question and was stuck after creating an equation. I think my approach was wrong.
Relevant Equations: ...

Here is my interpretation of the question. Three forces are applied to a bracket such that a 500N force acts at 30 degrees to the horizontal on the negative x-axis, while a 150N force acts at θ to the negative x-axis and another 150N force acts at 50+θ to the negative x-axis. The directions of the two 150N forces may vary, but the angle between the forces is always 50. Determine the range of theta values for which the magnitude of the resultant force acting at the bracket is less than 600N.

Per homework forum rules you must show us your attempt to receive help. We are to guide you to finding the solution, not hand it to you.

 

[Chestermiller]

Homework Statement: I've tried solving the question and was stuck after creating an equation. I think my approach was wrong.
Relevant Equations: ...

Here is my interpretation of the question. Three forces are applied to a bracket such that a 500N force acts at 30 degrees to the horizontal on the negative x-axis, while a 150N force acts at θ to the negative x-axis and another 150N force acts at 50+θ to the negative x-axis. The directions of the two 150N forces may vary, but the angle between the forces is always 50. Determine the range of theta values for which the magnitude of the resultant force acting at the bracket is less than 600N.

Let's see what you've done so far.

 

[CIA16]

Let's see what you've done so far.

I resolved the forces and came up with this equation. Here is where I got stuck.

 

[erobz]

View attachment 331056
I resolved the forces and came up with this equation. Here is where I got stuck.

Your force components seem to be consistent with $\rightarrow ^+$ and $\uparrow^+$: So at this point it's a math problem. Have you tried expanding and collecting like terms, and/or perhaps applying sum-difference formulas to the trigonometric entities?

 

[Lnewqban]

I see one little problem in order to exactly determine the highest value of the angular range in this question:

"Determine the range values of angle alpha for which the magnitude of the resultant of the forces acting at A is less than 600 N."

How far the two 150 N forces can be rotated CCW before the wall interferes with the nearest to it?

I have assumed that the limit position for that force is the vertical, resulting in 40 degrees, after a graphical scaled representation.

 

[CIA16]

I see one little problem in order to exactly determine the highest value of the angular range in this question:

"Determine the range values of angle alpha for which the magnitude of the resultant of the forces acting at A is less than 600 N."

How far the two 150 N forces can be rotated CCW before the wall interferes with the nearest to it?

I have assumed that the limit position for that force is the vertical, resulting in 40 degrees, after a graphical scaled representation.

I'll say we go along with your assumption.

 

[hmmm27]

I'll say we go along with your assumption.

Either that or make the arrows smaller so they don't bump the wall. Is the force of gravity involved in this at all ?

 

[CIA16]

Either that or make the arrows smaller so they don't bump the wall. Is the force of gravity involved in this at all ?

I don't think so. It's a statics problem that I think focuses solely on the resolution aspect.

 

[Lnewqban]

I'll say we go along with your assumption.

In that case, please see the attachment, which may help you simplify the equation you came up with in post #4.

 

[haruspex]

First, replace the two 150N forces with their resultant. You can find the magnitude and direction easily enough.
Second, get rid of the surd by squaring. Take a moment to think about what that does to the “<".
Third, expand the squares. You should get some simplification.
If you are still stuck, post what you get.

 

[CIA16]

In that case, please see the attachment, which may help you simplify the equation you came up with in post #4.

In your attachment, you got 27 degrees for alpha. How did you do that? Can you please show how you got the free-body diagram as well.

 

[haruspex]

In your attachment, you got 27 degrees for alpha. How did you do that? Can you please show how you got the free-body diagram as well.

Please try the plan in post #11.

 

[CIA16]

Please try the plan in post #11.

The resultant is 271.89N and the direction is 45°

 

[haruspex]

The resultant is 271.89N and the direction is 45°

Correct magnitude, wrong angle.

 

[CIA16]

Correct magnitude, wrong angle.

well, arctan(150/150) gives 45°. Is there something I'm missing? Quadrants?

 

[haruspex]

well, arctan(150/150) gives 45°. Is there something I'm missing? Quadrants?

That formula would be for two forces at right angles, one vertical, one horizontal.

 

[Chestermiller]

First, replace the two 150N forces with their resultant. You can find the magnitude and direction easily enough.
Second, get rid of the surd by squaring. Take a moment to think about what that does to the “<".
Third, expand the squares. You should get some simplification.
If you are still stuck, post what you get.

I like very much your idea of first getting the resultant of the two 150 N forces. This resultant is at an angle of $\alpha +25$ to the horizontal downward. Then the resultant can be parallel translated so that its tail is at the head of the 500 N force and it forms an angle of $180-(\alpha+55)$ degrees with the 150 N force. The resultant can then be determined using the law oof cosines.

 

[CIA16]

That formula would be for two forces at right angles, one vertical, one horizontal.

Yeah, I've seen my mistake. $$\theta = \tan^{-1}\left(\frac{F_2\sin(\theta)}{F_1 + F_2\cos(\theta)}\right)$$
gives $\theta$ to be equal to 25°. Then adding that to 30 for 500N and $\alpha$ is 55° + $\alpha$. If we find the resultant of that as well and introduce the inequality, after squaring both sides, we get $\alpha$ to be equal to 27.4°. But there's one more solution which should yield 223° and I'm not getting that.