Question

A weight$\mathit{W}$is supported by attaching it to a vertical uniform metal pole by a thin cord passing over a pulley having negligible mass and friction. The cord is attached to the pole$\mathbf{40.0}\mathbf{\text{\hspace{0.17em}cm}}$below the top and pulls horizontally on it (given figure). The pole is pivoted about a hinge at its base, is$\mathbf{1.75}\mathbf{\text{\hspace{0.17em}m}}$tall, and weighs$\mathbf{55.0}\mathbf{\text{\hspace{0.17em}N}}$. A thin wire connects the top of the pole to a vertical wall. The nail that holds this wire to the wall will pull out if an outward force greater than$\mathbf{22}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N}}$acts on it. (a) What is the greatest weight$\mathbf{\text{W}}$that can be supported this way without pulling out the nail? (b) What is the magnitude of the force that the hinge exerts on the pole

Short answer

1. The greatest weight$\mathrm{W}$that can be supported this way without pulling out the nail is$28.5\text{\hspace{0.17em}N}$.
2. The magnitude of the force that the hinge exerts on the pole $84.5\text{\hspace{0.17em}N}$

Step-by-step solution

Tension force

A tension force is a force created when a rope, string, or cable is stretched as a result of an external force. Along the length of the rope or cable, tension is exerted in the opposite direction as the force acting on it.

Identification of given data

Here we have given the distance from the top to the cord is $\mathrm{d}=40.0\text{\hspace{0.17em}m}$

The length of the pole is $\mathrm{L}=1.75\text{\hspace{0.17em}m}$

The weight of the pole is ${\mathrm{w}}_{\mathrm{p}}=55.0\text{\hspace{0.17em}N}$

The maximum outward force on the nail is ${\mathrm{F}}_{\max }=22.0\text{\hspace{0.17em}N}$

The angle between the wire and wall is$\mathrm{\theta }=37°$

Finding the greatest weight W that can be supported this way without pulling out the nail

(a)

Consider the following figure,

Here, the force exerted by the hinge, in general, has both horizontal and vertical components.

Let${\mathrm{F}}_{\mathrm{h},\mathrm{y}}$and ${\mathrm{F}}_{\mathrm{h},\mathrm{x}}$ be the vertical and horizontal force exerted by the hinge respectively.

We know that the horizontal component of the tension force equals to ${\mathrm{F}}_{\max }$.

So we have,

Now, substitute the values in the above equation. We get,

Now, the net torque about the hinge point has to be zero.

So, $(\mathrm{L}-\mathrm{d})\cdot \mathrm{W}-\mathrm{L}\cdot \mathrm{Tsin\theta }=0$

Hence, the greatest weight $W$that can be supported this way without pulling out the nail is $28.5\text{\hspace{0.17em}N}$.

Finding the magnitude of the force that the hinge exerts on the pole

(b)

Now, we want to find the magnitude of the force exerted by the hinge,${\mathrm{F}}_{\mathrm{h}}$.

By, the first condition of equilibrium, the net force must be zero both horizontally and vertically.

So we have,

$\begin{array}{c}-{\mathrm{w}}_{\mathrm{p}}-\mathrm{Tcos\theta }+{\mathrm{F}}_{\mathrm{h},\mathrm{y}}=0\\ {\mathrm{F}}_{\mathrm{h},\mathrm{y}}=55.0\text{\hspace{0.17em}N}+\left(36.6\text{\hspace{0.17em}N}\right)\cos 37.0°\\ {\mathrm{F}}_{\mathrm{h},\mathrm{y}}=84.2\text{\hspace{0.17em}N}\end{array}$

The magnitude of the hinge force is given by

$\begin{array}{c}{\mathrm{F}}_{\mathrm{h}}=\sqrt{{{\mathrm{F}}_{\mathrm{h},\mathrm{x}}}^2+{{\mathrm{F}}_{\mathrm{h},\mathrm{y}}}^2}\\ =\sqrt{{(-6.50\text{\hspace{0.17em}N})}^2+{\left(84.2\text{\hspace{0.17em}N}\right)}^2}\\ =84.5\text{\hspace{0.17em}N}\end{array}$

The magnitude of the force that the hinge exerts on the pole $84.5\text{\hspace{0.17em}N}$.