91Ó°ÊÓ

If an object's rotational velocity is constant, it is said to be in rotational equilibrium. A system must also have zero total torque in order to be in rotational equilibrium.

Here we have,$\mathrm{θ}={20.0}^{∘}$ .

The angle between rope and horizontal,$\mathrm{Î\pm }={58.0}^{∘}$

Slab width$\mathrm{a}=1.75\text{″¾}$

Slab height$\mathrm{b}=3.75\text{″¾}$

(a)

Consider the diagram for the given condition:

Now, force exerted by a single worker, $F_W$

So, by writing an equilibrium condition for rotation, the net torque about the lower right corner has to be zero:

Where, $w$is the weight of the slab.

Now, by substituting numerical values in above equation we get,

Hence, the force which each worker exert on the rope when$\mathrm{θ}={20.0}^{∘}$ is $0.011w$. Where $w$is the weight of the slab.

(b)

From part (a) we have,

$F_W=\frac{a\cos \mathrm{θ}−b\sin \mathrm{θ}}{10b\sin \mathrm{Î\pm }}â‹\dots w$

Here we can see that each worker has to exert less fore as$\mathrm{θ}$ increases.

As the weight's moment arm reduces by $\mathrm{θ}$, this makes sense. Because of this, the torque generated by the weight is reduced, which requires a smaller counter torque generated by the employees.

Hence, each worker has to exert less force.

(c)

The workers need to exert no force when the center of gravity of the slab lies directly above the support point.

So, the angle$\mathrm{θ}$ is at this moment determined by

Solve for the angle as:

This is the unstable equilibrium of the slab. If workers exert a small force at this moment, the slab will tip over.

The angle which the workers need to exert no force to balance the slab is $25.0°$.