How to find the angle and resultant force of three forces for a particle in equilibrium?

Question

A particle in equilibrium has three forces of magnitudes 5N, 6N and F N acting in it in the horizontal plane in the directions ( sorry it looks like I can’t attach a pic but the 5N is acting horizontally to the left, 6N diagonally right downwards with an acute angle of 70 degrees from the right hand horizontal and force F N acting diagonally upwards to the right with acute angle theta from the right hand horizontal, hope that makes sense :) ). Find the value of F and theta.

So been going over and over this but can’t come up with the correct answer of F = 6.36 (3.s.f) and theta = 70.5 degrees (1.d.p). It just gets quite confusing with two seperate angles in the same equilibrium and seems to get messy for me. Any help will be much appreciated:)

Answer 1

I'm going to approach this physics problem using the method of summing up forces.

Start by summing up the forces, or components of forces, in the x (horizontal) direction, and the y (vertical) direction.

In the positive x direction you have a component of the 6 N force, 6cos70°, and a component of the unknown force FcosΘ. In the negative x direction you have the 5 N force.

So, for the x direction: 6cos70° + FcosΘ = 5 rewrite this as FcosΘ = 5 - 6cos70°

In the positive y direction you have a component of the unknown force FsinΘ, and in In the negative y direction you have a component of the 6 N force, 6sin70°. The 5 N force has no y component.

So, for the y direction: FsinΘ = 6sin70°

You now have two equations and two unknowns. Combine them to solve for one of the unknowns. In this case it's easiest to find Θ first. FsinΘ / FcosΘ = 6sin70° / (5 - 6cos70°)

Cancel out the F's and you are left with tanΘ = 6sin70° / (5 - 6cos70°) giving you Θ = 62.4°, measured upward (counterclockwise) from the positive x axis.

Put this value of Θ into either of the x or y equations to get F = 6.36 N.

Answer 2

I am assuming that I understood your word description of the picture in a right way. My solution for the problem is like this:

Because the system is in equilibrium, the force F opposite to the resultant force F_{12 ,} where F₁₂ is the vector sum of force F₁ = 5 N and F₂ = 6 N. When you draw those 2 forces, you use the parallelogram rule to draw F₁₂. It will be a smallest diagonal. From the picture it is obvious that the magnitude of F₁₂ could be found with the use of the law of cosine.

We have

F₁₂ = √ (F₁² + F₂² - 2 F₁F₂ cos 70°) ,

or

F ₁₂ = √ (6² + 5² - 2·6·5·cos 70°) = 6.36 N

This force has the West of South direction. The angle Θ₁ between this F₁₂ force and horizontal we can find with the use of sine law.

We can write

F₁₂ / sin 70° = F₂ / sin Θ₁ .

From this equation

sin Θ₁ = F₂ · sin 70° / F₁₂

Putting the known values, we obtain

sin Θ₁ = 6 ·sin 70 ° / 6.36 = 0.8865 or Θ₁ = 62.4°

This angle, as we said, is formed by sides F₁ = 5 N and F₁₂ = 6.36 N.

Magnitude of the force F is the same as magnitude of F₁₂ . Hence, F = 6.36 N.

Angles Θ₁ and Θ are congruent, because they are vertical angles.

It means that the measure of angle Θ is 62.4º .

I believe that the answer for the angle is wrong. But just in case check my math again. Let me know, if you got different answer for angle.