M2~Rigid Body Equilibrium Assignment 2

百葉

2008-01-21 3:10 pm

A uniform strainght rod AB has length L and weight 2KW. The rod is in equilibrium with the end B in contact with a smooth vertical wall and the end A in contact with rough horizontal ground. The rod lies in a vertial plane perpendicular to the wall. The angle between the rod and the ground is Θ. A particle of weight W is attached to the rod at the point whose distance from A is XL. The magnitude of the force exerted the wall on the rod is √3 KW.

回答 (1)

2008-01-21 7:44 pm

✔ 最佳答案

圖片參考：http://i263.photobucket.com/albums/ii157/mathmate/math/Rigid2.jpg
Figure 1
Let
R=reaction from the wall at B (perpendicular to wall)
t=theta
Clockwise moment about A=RLsin(t)
Anti-Clockwise moment about A=W.XLcos(t)+2KW.(L/2)cos(t)
Equating moments about A:
W.XLcos(t)+2KW.(L/2)cos(t) = RLsin(t)
R=(X+K)Wcot(t)
sqrt(3)KW=(X+K)Wcot(t)
sqrt(3)K=(X+K)cot(t) ....(1)
The question does not explicitly define an unknown.
So equation (1) will be used if and when more data are available.

2008-01-22 12:04:24 補充：
1. Horizontal (Rh) and vertical (Rv) components of the reaction at A:Summation of horizontal forces = R Rh=0Rh=-R=-sqrt(3)KW (towards wall)Summation of vertical forces=Rv 2KW W=0Rv=-(2K 1)W (upwards)

2008-01-22 12:06:37 補充：
From equation (1),X=K(sqrt(3)-cot(theta))/cot(theta)for X =0, sqrt(3) =cot(theta)or cot(theta) =30degrees

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