Equilibrium of Rigid Bodies Problem

[Mugwump101]

Homework Statement

A hand truck is used to move two barrels, each weighing 80 lb.
Neglecting the weight of the hand truck, determine (a) the vertical force P
which should be applied to the handle to maintain equilibrium when
α = 35°, (b) the corresponding reaction at each of the two wheels.

Link for diagram: http://people.clarkson.edu/~hhshen/Courses%20Page/Courses/Course6/teach/review2.pdf

Go to prob. 4.5!

Homework Equations

Sum of all forces = 0
Sum of all moments = 0

The Attempt at a Solution

I honestly just don't understand this period. As far as I got was to make a big triangle where it's 64sin35 to find the side on the ground which is 52.42 which would give us the M= rxF for P. 52.42xP at Mb. Ohhhh, would I have to do that for the forces at g2 and g1 (to find the sides on the 'ground' and cross it with the weights to find P in the equation of the sum of the moments? If so, what about for part b?

Thanks a lot in advance!

 

[PhanthomJay]

There are 2 ways to calculate moments. One is M=rXF =rFsin theta , where r is the magnitude of the vector between the point of application of F and the pivot point, and theta is the angle between the r and F vectors. The other method, preferred by most engineers, is Moment = Force times perpendicular distance from line of action of the force to the pivot point. Using the first method and looking at the moment about B from P, then M = P r sin (90-alpha). Using the 2nd method, M = P r cos alpha. Same result. Watch signs (cw vs ccw). Sum moments of all forces =0. The solution manual explains it very well. For part b, sum of forces in y direction =0 (there are 2 wheels).

 

[pongo38]

Can I add that I think your horizontal distance, though correct, was not 64sin35 but 64cos35. This brings out the need for you to draw it to scale and measure the result as a check that your sin and cos are not muddled (as they often are).

 

[mgier001]

I know this thread is, well... ancient. But I have this problem assigned for homework, and I understand all about moment and that type of thing. My problem is that I don't understand how a1 and a2 are derived in the solutions manual.
a1=20sin(alpha)-8cos(alpha)
a2=32cos(alpha)-20sin(alpha)
That's pretty much my big problem :\, I'd appreciate all help!
Thanks very much!
-Matt

 

[PhanthomJay]

a1 and a2 are found using geometry and trig. You have to look at both the problem sketch and the fbd to see how they are calculated. It isn't that easy without the solution sketch. Study it carefully.

 

[mgier001]

Unfortunately my geometry seems to be lacking, and that's why I'm no understanding how to get the dimensions. Also, I spent a great amount of time already studying the solutions drawing and still couldn't understand :(. Do you mind taking me step by step showing how the sin and cos are used for a1? I understood the 8cosalpha, but I don't get why they used sin also.

 

[PhanthomJay]

The distance from G to B is 20 in. Measured perp to the incline. Now note the angle alpha at B. the 20 is the hypotenuse of a right triangle. 20 sin alpha is the horizontal leg.