题目
Problem
2.
A plank AB, of length 6m and mass 4kg, rests in equilibrium horizontally on two supports at C and D, where AC=2m and DB=1m.
A brick of mass 2kg rests on the plank at A and a brick of mass 3kg rests on the plank at B, as shown in Figure 1.
The plank is modelled as a uniform rod and all bricks are modelled as particles.
(a) Find the magnitude of the reaction exerted on the plank
(i) by the support at C,
(ii) by the support at D.
(6)
The 3kg brick is now removed and replaced with a brick of mass xkg at B.
The plank remains horizontal and in equilibrium but the reactions on the plank at C and at D now have equal magnitude.
(b) Find the value of x.
(4)
题目中文翻译
一块长 6m、质量为 4kg 的木板 AB 水平静止平衡于 C、D 两点的支座上,其中 AC=2m,DB=1m。
如图 1 所示,一块质量为 2kg 的砖放在木板的 A 点,一块质量为 3kg 的砖放在木板的 B 点。
将木板视为均匀杆,将所有砖视为质点。
(a) 求下列支座对木板施加的支持力大小:
(i) C 点的支座,
(ii) D 点的支座。
(6)
现移去 3kg 的砖,换成一块质量为 xkg 的砖放在 B 点。
木板仍保持水平平衡,但此时 C、D 两点处对木板的支持力大小相等。
(b) 求 x 的值。
(4)
解答
(a)
解法一
思路
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均匀木板的重心在中点。先绕 D 取矩以消去 D 的支持力,再用竖直平衡求另一支持力。
答题过程
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Let the reactions at C and D be RC and RD. The plank’s weight 4g acts at its midpoint G, with
CG=1 m,GD=2 m,CD=3 m.
(i) Taking moments about D,
3RC+(3g)(1)=3RC=RC=(4g)(2)+(2g)(5),15g,5g=49 N.
(ii) Resolving vertically,
RC+RD=(4+2+3)g=9g.
Thus
RD=9g−5g=4g=39.2 N.
(b)
解法一
思路
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换砖后支持力需要重新计算。将两个新支持力共同记为 R,结合总重量和力矩方程消去 R。
答题过程
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Let each new reaction be R. Vertical equilibrium gives
2R=(6+x)g.
Taking moments about D,
3R+xg=4g(2)+2g(5)=18g.
Substituting R=(6+x)g/2 and cancelling g,
23(6+x)+x=9+25x=x=18,18,3.6.