Engineering Mechanics - Equilibrium of a Rigid Body - Discussion

Discussion Forum : Equilibrium of a Rigid Body - General Questions (Q.No. 1)
1.

The girl has a mass of 17kg and mass center at Gg, and the tricycle has a mass of 10kg and mass center at Gt. Determine the normal reactions at each wheel for equilibrium.

NA = 14.77 N, NB = NC = 6.12 N
NA = 128.8 N, NB = NC = 68.0 N
NA = 144.9 N, NB = NC = 60.0 N
NA = 13.15 N, NB = NC = 6.93 N
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.

JBM SAURABH SHARMA said:   3 years ago
Just take moment about A;

You get (17*9.81*450) +(10*9.81*450) = 2RB*750
As RB = RC at mid, it will be 2RB.

From here you will get RB = RC = 60N,
Now take moment about 2RB.

You will get (300*10*9.81) +(475*17*9.81) = RA*750.
From here RA = 144.9 N.

Hence option C is the Correct answer.

Sreerag said:   1 decade ago
Please answer the question that 3 cylinders are placed in a rectangular channel or shown in the figure. Determine the reaction between the cylinder 1 and vertical wall.

Sam said:   1 decade ago
Force at Gg = 17g N where g=9.81
Force at Gt = 10g N

Distance Gg -> A = 450
Distance Gg -> CD = 300

Force A = 300/750 * 17g = 66.6 N
Force CD = 450/750 * 17g = 100N

Distance Gt -> A = 275
Distance Gt -> CD = 425

Force A = 425/700 * 10g = 59.5N
Force CD = 275/700 * 10g = 38.5N

Total force A = 59.5+66.4 = 126 N (which is close approx.)
Total force CD = 100+38.5 = 138.5N
Force at each wheel = 138.5/2 = 69N (approx.)

B is the closest answer
(1)

Rangi said:   1 decade ago
please answer

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