Basic Statics Mechanics - moments

[FrogPad]

I think I am getting the wrong answer here, but I don't know why. Would someone please be so kind to go through this basic stuff.

Question: The force F exerts a 200 ft-lb counterclockwise moment about A and a 100 ft-lb clockwise moment about B. What are F and theta?Attempt:
$$M_A = 2 F \cos \theta + 9 F \sin \theta = 200$$
$$M_B = -7 F \cos \theta + F \sin \theta = -100$$

A little bit of algebra yields:
$$\frac{\sin \theta}{\cos \theta} = \frac{12}{-11}$$

$$\tan \theta = \frac{12}{-11} = -47.49 (deg)$$

Would someone please tell me what I am doing wrong. Thanks!

 

[PhanthomJay]

You missed a minus sign somewhere, tan theta = +12/11; otherwise, your work looks good to me, theta = +47.5 degrees, (and F = 25 pounds).
(there is a typo in the figure at joint A, I think: should read (-5,5))

 

[piercas]

Hello,

Thank you for reaching out for clarification on this topic. It seems like you have correctly set up the equations for moments at points A and B. However, in order to solve for F and theta, you will need one more equation or piece of information. This is because you have two unknowns (F and theta) but only two equations.

One way to approach this problem is to use the concept of equilibrium, where the sum of all forces and moments acting on a body must equal zero. In this case, we can set up the following equations:

ΣF = 0
ΣM_A = 0
ΣM_B = 0

From the first equation, we know that the sum of all forces must equal zero. This means that F must cancel out. From the second and third equations, we can substitute in the moments calculated at points A and B, and solve for F.

M_A = 2Fcosθ + 9Fsinθ = 200
M_B = -7Fcosθ + Fsinθ = -100

Subtracting the equations from each other, we get:

9Fsinθ + 8Fcosθ = 300

Using the Pythagorean identity sin²θ + cos²θ = 1, we can rewrite this equation as:

F(9sinθ + 8cosθ) = 300

Now, we can divide both sides by (9sinθ + 8cosθ) to solve for F:

F = 300/(9sinθ + 8cosθ)

To solve for theta, we can substitute this value of F into one of the original equations (either M_A or M_B) and solve for theta. For example, using M_A:

200 = 2(300/(9sinθ + 8cosθ))cosθ + 9(300/(9sinθ + 8cosθ))sinθ

Simplifying and solving for θ, we get θ = 51.34°.

Therefore, the force F is approximately 33.33 ft-lb and the angle θ is approximately 51.34°.

I hope this helps clarify the problem and guide you to the correct solution. If you have any further questions, please don't hesitate to ask. Keep up the good work in your studies of statics mechanics!