How do I solve for B in this moment and vector homework?

[ThomasHW]

Homework Statement

http://www.tunerspec.ca/school/question_oct16.jpg

Homework Equations

M=FD

The Attempt at a Solution

http://www.tunerspec.ca/school/work_oct16.jpg

I'm having trouble solving for B by adding the moments about the point A. I am getting an answer of 4.31kN for the force of B (as you can see on the bottom of my work,) but this is wrong.

Can someone please help?

 

[Pyrrhus]

Your sign convention for the moments has a contradiction when you say 32 kn force horizontal component's moment is negative with respect to A, and then the y component's moment is positive with respect to A, when both are counterclockwise.

 

[ThomasHW]

Isn't the y-component trying to rotate it counterclockwise, while the x-component is trying to rotate it clockwise?

 

[Pyrrhus]

Isn't the y-component trying to rotate it counterclockwise, while the x-component is trying to rotate it clockwise?

Both are rotating the body counterclockwise if A is the fixed point, or where the axis of rotation goes through. Imagine the body rotating about A, and the influence of the components.

Another way to see this is by projecting the line of action of the x component of the force until it intersects the body. (it'll be 200 mm under A)

 

[ThomasHW]

Ahh, okay. So I have the 34kN sign convention correct? It's kind of confusing deciding which way you think it'll go.

 

[ThomasHW]

Another question... is B negative? If so, can you explain why?

 

[Pyrrhus]

Ahh, okay. So I have the 34kN sign convention correct? It's kind of confusing deciding which way you think it'll go.

Yes is correct.

It can be confusing, but if you always PROJECT the line of action of the forces, you will know. Remember the transmissibility principle of Statics, a force can move along its line of actions without changing its external effects over a rigid body.

 

[Pyrrhus]

Another question... is B negative? If so, can you explain why?

What do you mean by B negative? its moment about A?

 

[ThomasHW]

What does projecting the line of action of the force tell me about the sign convention?

I'm working along a problem in my textbook which is exactly the same, except the angles and forces are SLIGHTLY different. It has Ay and B as being negative, and Ax as being positive. I figured out how Ay is negative (in my work) but I'm not seeing how Ax is positive if B is negative (Using the Fx equation at the top of the page in my work.)

 

[Pyrrhus]

What does projecting the line of action of the force tell me about the sign convention?

I'm working along a problem in my textbook which is exactly the same, except the angles and forces are SLIGHTLY different. It has Ay and B as being negative, and Ax as being positive. I figured out how Ay is negative (in my work) but I'm not seeing how Ax is positive if B is negative (Using the Fx equation at the top of the page in my work.)

The truth is there's not such thing as "negative" in vectors, because forces are vectors and their magnitude (modulus) is always positive. Negative just means the force (vector) is in the "wrong" direction.

An example will be is your posted problem, in order to guarantee a null resultant force, the Ay vector must point downwards.

 

[ThomasHW]

Thank you for the help! :)