phy-force - Ans.fyi 參考答案

phy-force

2011-07-25 3:15 am

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A uniformrod AB, of length 4m and mass 6kg, rests in equilibrium with one end, A, onsmooth horizontal ground. The rod rests on a rough horizontal peg at the pointC, where AC is 3m. The rod is inclined at an angle of 20° to the horizontal.a) find the magnitude of thenormal reaction force between the rod and the ground.bi) find the normal reactionacting on the rod at C.bii) find the friction forceacting on the rod at C

回答 (1)

天同

2011-07-25 5:51 am

✔ 最佳答案

(a) Take moment about CRa(3cos(20)) = 6g.cos(20)where Ra is the normal reation at A and g is the acceleration due to gravity (taken to be 10 m/s2)hence, Ra = 60/3 N = 20 N
(b) Let Rc be the normal reaction at C and Ff be the friction at C
In the vertical direction,Rc.cos(20) + Ra + Ffsin(20) = 6g
i.e. Rc.cos(20) + 20 + Ff.sin(20) = 60
Rc.cos(20) + Ff.sin(20) = 40 -------------------- (1)

In the horizontal direction,
Rc.sin(20) = Ff.cos(20)
thus, Ff = Rc.sin(20)/cos(20) ----------------- (2)

equation (1) becomes,
Rc.cos(20) + Rc.[sin(20)]^2/cos(20) = 40
solve for Rc gives Rc = 37.6 N

(c) From equation (2)
Ff = 37.6.sin(20)/cos(20) N = 13.7 N

2011-07-25 20:49:18 補充：
In b(i), you can take moment about A
Rc.x 3 = 6g x 2cos(20).
hence, Rc = 37.6 N
This is much simpler.

2011-07-25 20:50:27 補充：
Then use equation (2) to find out Ff in b(ii)

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