# Statics problem on the resolution of forces

• CIA16
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:
$$\sqrt{(-500 \cos(30) - 150 \cos(x) - 150 \cos(50 + x))^2 + (500 \sin(30) - 150 \sin(x) - 150 \sin(50 + x))^2}<600$$

#### Attachments

• photo_2023-08-25_11-44-50.jpg
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CIA16 said:
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.

CIA16 said:
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.

Chestermiller said:
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.

CIA16 said:
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?

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Chestermiller
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.

Lnewqban said:
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.

CIA16 said:
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 ?

hmmm27 said:
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.

CIA16 said:
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.

#### Attachments

• Resultant force.pdf
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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.

Chestermiller
Lnewqban said:
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.

CIA16 said:
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.

haruspex said:
Please try the plan in post #11.
The resultant is 271.89N and the direction is 45°

CIA16 said:
The resultant is 271.89N and the direction is 45°
Correct magnitude, wrong angle.

haruspex said:
Correct magnitude, wrong angle.
well, arctan(150/150) gives 45°. Is there something I'm missing? Quadrants?

CIA16 said:
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.

haruspex said:
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.

haruspex said:
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.

## What is the resolution of forces in statics?

The resolution of forces in statics involves breaking down a single force into two or more components that collectively have the same effect as the original force. This is typically done using vector decomposition into perpendicular components, such as horizontal and vertical forces.

## Why is resolving forces important in statics problems?

Resolving forces is crucial because it simplifies the analysis of complex systems. By breaking down forces into their components, we can apply the principles of equilibrium to each direction independently, making it easier to solve for unknowns and ensure the system is in a state of balance.

## How do you resolve a force into its components?

To resolve a force into its components, you typically use trigonometric functions. If you know the magnitude of the force and the angle it makes with a reference axis, you can use the sine and cosine functions to find the perpendicular components. For example, a force F at an angle θ can be resolved into F*cos(θ) in the horizontal direction and F*sin(θ) in the vertical direction.

## What are the common mistakes to avoid when resolving forces?

Common mistakes include incorrectly identifying the angle, using the wrong trigonometric function, and not maintaining consistent units. Another frequent error is neglecting to consider all forces acting on the body, which can lead to incorrect conclusions about the system's equilibrium.

## Can you give an example of a statics problem involving the resolution of forces?

Consider a beam supported at one end and subjected to a force at an angle. To find the reactions at the support, you would resolve the applied force into horizontal and vertical components. Then, you apply the equilibrium equations (sum of forces in horizontal and vertical directions and the sum of moments about a point) to solve for the unknown reactions. This process helps ensure that the beam remains in static equilibrium.

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