Question

Two people carry a heavy object by placing it on a uniform board that is 2.00m long and weighs 2.00x102 N. One person lifts at one end with a force of 4.00x102 N, and the other lifts the opposites end with a force of 6.00x102 N. Find the weight and the location of the heavy object. 

Answer 1

Explanations & Calculations

• Let's assume the system stays in equilibrium without collapsing.
• Then the forces, as well as the torques, are balanced out & that equilibrium could be considered in solving this.

• Equilibrium of forces,
• And if the weight they are lifting is $\small W$

$\qquad\qquad \begin{aligned} \small \Sigma F&=\small \\ \small4.00\times10^2+6.00\times10^2-2.00\times10^2-W&=\small 0\\ \small W&=\small \bold{8.00\times10^2\,N} \end{aligned}$

• Since the board is uniform, its center of gravity lies on the geometrical center which is at 1.00m from either person.

• Equilibrium of torques,
• If its point of action lies at some $\small x$ distance from the person's end who applies $\small 4.00\times10^2\,N$.

$\qquad\qquad \begin{aligned} \small \curvearrowright\Sigma \tau&=\small 0\\ \small W.x+(2.00\times10^2N\times 1.00m)-(6.00\times10^2N\times2.00m) &=\small 0\\ \small x&=\small \frac{1000Nm}{8.00\times10^2N}\\ &=\small \bold{1.25\,m} \end{aligned}$