Force Resultant Along x' Axis: Find F & Theta

[Robb]

Homework Statement

Three forces act on the bracket, the resultant force is directed along the x' axis and has a magnitude of 9.8kN. Determine the magnitude of F. Determine the direction theta of F.

Homework Equations

The Attempt at a Solution

I'm really not sure how I am doing with. Any advice would be much appreciated!

[/B]

 

[gneill]

Robb,

You'll get more responses and better help if helpers can see clear and commented work that's easy to quote in a response. A not-too-clear image of handwritten calculations without comments will deter many. It is better to insert a description of what you are doing and why at each step. The very best thing to do, of course, is to type out your math using LaTeX so that helpers can directly quote and comment...

 

[kuruman]

In your drawing (that is hard to read) I see that you have attempted to write the forces using unit vector notation. That is good. However, these forces are in two dimensions in the plane of the screen. You seem to have written the forces in three dimensions. Write all three forces in unit vector notation using ## \hat{i} ## and ## \hat{j} ## only, add them and see what is needed to get a resultant in the direction x'.

On edit: Try to follow gneill's recommendation.

 

[Robb]

Sorry for the confusing drawing. I guess I was looking at x' as a third axis/dimension. No?

 

[kuruman]

No, x' is in the same plane as the 4kN, the 6kN and the resultant. It has to be in the same plane because it is the direction of the resultant. Maybe this will make it easier for you to find the answer.

 

[Robb]

F1= 6i
F2= F(2x)i + F(2y)j
F3= (-4sin15)i + ( 4cos15)j= -1.03i = 3.86j
Fr- (9.8cos30)i +(9.8sin30)j = 8.49i + 4.9j

8.49= 6-1.03 + F(2x)
F(2x)= 3.52kNi
F(2y)= 4.9 - 3.86= 1.04kNj

F2= 3.67kN

I assume for the direction theta, they are referring to the one shown between the y-axis and the vector F2? I labeled the vectors F1, F2, F3 starting with the x-axis and moving counter clockwise.

 

[kuruman]

assume for the direction theta, they are referring to the one shown between the y-axis and the vector F2?

Yes, that's what they mean. Find it and you are done. I didn't run your numbers but they look OK.

 

[Robb]

arctan(1.04/3.52) = 16.46

Looking at the drawing this angle would make sense but which angle is tangent measuring here? Typically I would say between the x-axis and F2 but that number doesn't make sense looking at the drawing.

 

[kuruman]

Drop a perpendicular from the tip of F to the y-axis. You have a right triangle and the perpendicular you just drew is opposite to θ. What is tanθ?

 

[Robb]

arctan (1.04/3.52)= 16.46

I looked at it that way to but I also wondered about dropping a perpendicular from the tip of F to the x axis. NVMD! Just saw it. Thanks for the help!

 

[Robb]

So, the answer turned out to be theta= 73.5. Seems to me, from looking at the drawing, like that makes more sense if it were measured from the x axis, ccw.

 

[kuruman]

The drawing shows that θ is measured with respect to the y-axis CW. You have to find the angle that the problem is asking you to find and 73.5^o is the complementary angle of θ. Not all angles are measured with respect to the x-axis CCW.

 

[Robb]

That's what I though but masteringengineering.com says the answer for theta is 73.5. I entered 16.5 first and it said I was wrong. I really hate that site for homework!

 

[kuruman]

I would hate it too. You have learned all you need to learn from this problem. I suggest that you show your solution to your instructor and ask why it is marked wrong. He/She needs to know and, if enough people complain, perhaps a different homework platform will be chosen next time the course is taught.