 Aug 3rd 2018, 05:13 AM   #1
Junior Member

Join Date: Jul 2018
Posts: 2
Free body diagram

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.
Attached Files figures.pdf (113.3 KB, 20 views)   Aug 7th 2018, 05:07 AM #2 Senior Member   Join Date: Jun 2016 Location: England Posts: 929 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 topsquark likes this. __________________ ~\o/~ Last edited by Woody; Aug 7th 2018 at 05:13 AM.   Aug 8th 2018, 03:33 AM #3 Junior Member   Join Date: Jul 2018 Posts: 2 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.   Aug 9th 2018, 05:46 AM #4 Senior Member   Join Date: Jun 2016 Location: England Posts: 929 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) __________________ ~\o/~   Aug 9th 2018, 08:18 AM #5 Senior Member   Join Date: Jun 2010 Location: NC Posts: 417 Glorian, All... I'm unable to display the attached "pdf." ??? Any trick to thiis? JP   Aug 9th 2018, 08:30 AM #6 Senior Member   Join Date: Jun 2016 Location: England Posts: 929 opening pdf I just did a "right click" and selected "open in new window" from the option list. __________________ ~\o/~  Tags body, diagram, free Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Steelers72 Kinematics and Dynamics 3 Feb 28th 2015 05:23 AM ManyArrows Equilibrium and Elasticity 1 Apr 26th 2009 12:28 AM Apprentice123 Kinematics and Dynamics 0 Apr 7th 2009 12:15 PM Apprentice123 Kinematics and Dynamics 0 Apr 4th 2009 10:48 AM Apprentice123 Kinematics and Dynamics 2 Apr 4th 2009 07:38 AM 