Question

6 s figure 4.22 shows a uniform metal bar suspended by ropes at points p and q
a spotlight is suspended from the bar. the weight w of the bar and the weight l of the spotlight are 360 n and 150
respectively. the forces in the ropes at points p and q are x and y respectively. the bar is in equilibrium.

figure 4.22
(a) state, in terms of x, the moment of force x about point q
(b) calculate the following and give your answers to 3 significant figures.
(i) the moment of w about point q
moment = _ n cm
(ii) the moment of l about point q
moment = _ n cm
(iii) force x =
force x = _ n
(iv) force y
force y = _ n

Explanation:

Step1: Moment of force formula

The moment of a force \(M = F\times d\), where \(F\) is the force and \(d\) is the perpendicular distance from the pivot.

Step2: (i) Moment of \(W\) about \(Q\)

Given \(W = 360\space N\) and \(d=70\space cm\). Using \(M = F\times d\), we have \(M_W=360\times70 = 25200\space N\cdot cm\)

Step3: (ii) Moment of \(L\) about \(Q\)

The distance of \(L\) from \(Q\) is \(70 + 40=110\space cm\). Given \(L = 150\space N\). Using \(M = F\times d\), we have \(M_L=150\times110=16500\space N\cdot cm\)

Step4: (iii) Force \(X\) (using principle of moments about \(Q\))

For the bar in equilibrium, \(\sum M_{Q}=0\). Taking clock - wise moments equal to anti - clockwise moments. The anti - clockwise moment is due to \(X\) and clock - wise moments are due to \(W\) and \(L\). So \(X\times140=360\times70 + 150\times110\) (since the distance of \(X\) from \(Q\) is \(70 + 70=140\space cm\)).

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Step5: (iv) Force \(Y\) (using vertical equilibrium \(\sum F_y = 0\))

Since the bar is in vertical equilibrium \(X + Y=W + L\). We know \(X = 298\space N\), \(W = 360\space N\) and \(L = 150\space N\).

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Answer:

(i) \(25200\space N\cdot cm\)
(ii) \(16500\space N\cdot cm\)
(iii) \(298\space N\)
(iv) \(212\space N\)