Question

URGENT----PLEASE SOLVE !!

Answer 1

$\setlength{\unitlength}{1mm}\begin {picture}(5,5)\multiput(0,0)(0,0.5){2}{\line(1,0){50}}\multiput(0,0)(50,0){2}{\line(0,1){0.5}}\multiput(0,0)(50,0){2}{\line(0,-1){10}}\multiput(-2,-14)(50,0){2}{\framebox(4,4)}\multiput(0,-14)(50,0){2}{\vector(0,-1){10}}\put(2,-24){39.2 N}\put(52,-24){58.8 N}\put(29,-2.5){ \triangle } \put(25,0){\vector(0,-1){10}}\put(27,-10){39.2 N}\put(12.5,-2){\vector(-1,0){12.5}}\put(16.5,-2){\vector(1,0){12.5}}\put(13.5,-2.5){x}\end {picture}$

Here the weights of 4 kgf = 4 × 9.8 = 39.2 N and 6 kgf = 58.8 N are suspended at the ends of a uniform rod of length 0.5 m and weight 4 kgf = 39.2 N.

Let a peg be pointed on the rod at a distance x metres from the end where the weight 39.2 N is suspended to bring the rod in equilibrium. So we can say that this point, where the peg is pointed, is the centre of weight of the rod.

• The weight 39.2 N is suspended at x metres away from the centre of weight.
• The weight of the rod 39.2 N is acting at its centre. Since the rod is 0.5 m long, the weight of the rod is acting at 0.25 m away from any of its end. So we can say that its weight is acting at a distance (x - 0.25) metres away from the centre of weight towards left.
• The weight 58.8 N is suspended at (0.5 - x) metres away from the centre of weight.

We know that the net torque acting on any body is zero at the centre of weight. Then we can have the equation, at the centre of weight,

$\mathrm{39.2x+39.2(x-0.25)-58.8(0.5-x)=0}$

where the downward forces at left of centre of weight are taken as positive.

So,

$\mathrm{39.2x+39.2x-39.2\times0.25-58.8\times0.5+ 58.8x=0}\\\\\\\mathrm{137.2x-39.2=0}\\\\\\\mathrm{x=\dfrac{39.2}{137.2}}\\\\\\\boxed{\mathbf{x=0.29\ m}}$

Hence the peg should be pointed on the rod at a distance around 0.29 m away from the end where the weight 39.2 N is suspended.

#answerwithquality

#BAL