Free body diagram

Jul 2018
3
0
Greetings!

I could need some help on the following problem.

1. Explain force and momentum equilibrium for a free body and the division of forces in composants. Show how these are used to determine the forces on the two rods in the figure.

2. Explain what tension there are in the rods and the bolts holding the rods to point A and C.


For part 1 have I attemped the following:

∑Fx = 0: Ax – T cos 60 = 0

∑Fy = 0: Ay + sin 60 – 75 = 0

∑Mx = 0: T sin 60 - 75 = 0


Rod AB.

75 = T sin 60

T = 75 / sin 60

T = 86,6 N


Ax = T cos 60

Ax = 86,6 cos 60

Ax = 43,3 N


Ay = 0


Rod AB is pivoted at points A and B and since it is horizontal, it can only exert force in the horizontal direction. So the vertical force exerted by rod AB on pivot A is zero.


Rod BC.

75 = T sin 60

T = 75 / sin 60

T = 86,6 N


Bx = T cos 60

Bx = 86,6 cos 60

Bx = 43,3 N


By + sin 60 – 75 = 0

By = - sin 60 + 75

By = 74,3 N.
 

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Jun 2016
1,150
520
England
I think your first point of thought should be:
F1+F2+F3=0

This implies that the individual x & y components can be equated similarly:
F1x+F2x+F3x=0
and:
F1y+F2y+F3y=0

This seems to be the sort of direction you are trying to go in,
but I think you are trying to jump to the individual tensions too soon.

I think a major hint might be the size of F1y
and similarly the magnitude of F2x
 
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Jul 2018
3
0
Hi!

Thanks for the comment. I have attemped the following:

The net force on pivot B is:

0 = F1 + F2 + F3.


Rod AB.
∑Fx = 0: Ax – T cos 60 = 0

∑Fy = 0: Ay + sin 60 – 75 = 0

∑Mx = 0: T sin 60 - 75 = 0


∑Mx = 0:

75 = T sin 60

T = 75 / sin 60

T = 86,6 N


∑Fx = 0:

Ax = T cos 60

Ax = 86,6 cos 60

Ax = 43,3 N


∑Fy = 0:

Ay = 0



Rod CB.

∑Fx = 0: Cx - T cos 60 = 0

∑Fy = 0: Cy - sin 60 – 75 = 0

∑My = 0: T sin 60 - 75 = 0

∑My = 0:

75 = T sin 60

T = 75 / sin 60

T = 86,6 N


Cx = T cos 60

Cx = 86,6 cos 60

Cx = 43,3 N


Cy - sin 60 – 75 = 0

Cy = sin 60 + 75

Cy = 75,9 N.


Rod AB is pivoted at points A and B and since it is horizontal, it can only exert force in the horizontal direction. So the vertical force exerted by rod AB on pivot A is zero.

Rod CB is pivoted at points B and C and since it is at a 60 deg angle to the horizontal and the vertical, it exert a horizontal force of -75 N, and a vertical force on pivot C of 75 kN.
 
Jun 2016
1,150
520
England
Your approach is somewhat different to mine,
but I think you are on the right lines.

My thinking is:

F1 is horizontal so F1y = 0 thus:
F3y=-F2y
but we know that F2y = 75kN so:
F3y=-75

F2 is vertical so F2x = 0 thus:
F1x=-F3x

Lastly we know the F3 is at 30deg to the vertical so:
F3x/F3y=tan(30)
 
Jun 2010
422
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NC
Glorian, All...

I'm unable to display the attached "pdf." ???

Any trick to thiis?

JP
 
Jun 2016
1,150
520
England
opening pdf

I just did a "right click" and selected "open in new window" from the option list.