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.
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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