Chapter 8, Problem 30P

(a)

To determine

Draw a force diagram and their placement on the strut.

Explanation of Solution

The weight of the beam acting downward from its midpoint, tension of the acting left towards the wall and the horizontal force of the wall on the strut acts towards right and the vertical force of the wall on the strut acts upward.

The Free body diagram for force acting on the strut is given below.

Conclusion:

The free body diagram of the forces and their position is drawn.

(b)

To determine

Whether the hinge is a good place to use for calculating torques.

Explanation of Solution

The point about which the object rotates is known as the axis of rotation. Torque depends on the force and the distance where the force acts from the hinge.

The force due to the hinge is unknown force and these forces are not included in the torque calculation because the distance where the force acts from the hinge is zero. Therefore the torque at one end is equal to zero. So that to simplify the torque equation is easy.

Conclusion:

The hinge is a good place to use for calculating torques because torque at one end is zero, so it is easier to solve the resulting equilibrium equation.

(c)

To determine

The condition for rotational equilibrium to calculate torque around the hinge.

Answer to Problem 30P

The condition for rotational equilibrium to calculate torque around the hinge is $\left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]=\left[T\left(L\sin \text{θ}\right)\right]$

Explanation of Solution

Given info: length of the strutis $3.00\text{m}$, mass of the strutis $16.0\text{kg}$ and the angle strut makes with the cable is $\text{30}\text{.0}°$.

Formula for the torque about the hinge is,

$\sum {\tau }_{\text{hinge}}=\left[\left(mg\right)\left(\frac{L}{2}\cos \theta \right)\right]-\left[T\left(L\sin \theta \right)\right]$

• • $\sum {\tau }_{\text{hinge}}$ is the net torque about the hinge
  • L is the length of the strut
  • m is the mass of the strut
  • g is the acceleration due to gravity
  • T is the tension in the cable

The system is in equilibrium, So, the condition for rotational equilibrium is the outward torque is equal to the inward torque.

The condition for rotational equilibrium to calculate torque around the hinge is,

$\sum {\tau }_{\text{hinge}}=0$

Thus, the formula for the torque about the hinge is,

$\begin{array}{c}0=\left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]-\left[T\left(L\sin \text{θ}\right)\right]\\ \left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]=\left[T\left(L\sin \text{θ}\right)\right]\end{array}$

Conclusion:

The condition for rotational equilibrium to calculate torque around the hinge is $\left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]=\left[T\left(L\sin \text{θ}\right)\right]$.

(d)

To determine

Tension in the cable byUsing torque equation.

Answer to Problem 30P

The tension in the cable is $132\text{N}$.

Explanation of Solution

Given info:The length of the strut L is $3.00\text{m}$ , mass of the strut is $16.0\text{kg}$, acceleration due to gravity g is $9.8\text{m}/{\text{s}}^{\text{2}}$ and an angle $\theta$ is $\text{30}{\text{.0}}^{\text{o}}$

Formula to calculate the tension in the cable is,

$\left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]=\left[T\left(L\sin \text{θ}\right)\right]$

Rearrange in terms of T.

$\begin{array}{c}T=\frac{\left[\left(mg\right)\left(\frac{L}{2}\cos \text{θ}\right)\right]}{\left[\left(L\sin \text{θ}\right)\right]}\\ =\frac{\left(mg\right)}{2\tan \text{θ}}\end{array}$

Substitute $16.0\text{kg}$ for m, $\text{30}\text{.0}°$ for $\theta$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for acceleration due to gravity g to find T.

$\begin{array}{c}T=\frac{\left(16.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{2\left(\tan \text{30}\text{.0}°\right)}\\ =132\text{N}\end{array}$

Conclusion:

The tension $\left(T\right)$ in the cable is $132\text{N}$.

(e)

To determine

The x- and y-components of Newton's second law for equilibrium.

Answer to Problem 30P

The x and y-components of Newton's second law for equilibrium $F_x$ is $T$ and $F_y$ is $mg$.

Explanation of Solution

Given info:

Formula to calculate the x- components of tension from Newton's second law for equilibrium is,

$F_{\text{net},x}=F_x-T$

• $F_{\text{net},x}$ is the net horizontal force acting
• $F_x$ is the horizontal force on the strut from the wall
• T is the tension in the cable

Since the strut is in equilibrium $F_{\text{net},x}=0$.

Formula to calculate the x- components of tension from Newton's second law for equilibrium is,

$\begin{array}{c}{F}_x-T=0\\ F_x=T\end{array}$

Tension in the cable acts left towards the wall to counter the tension, the wall exerts a force acting toward right.

Formula to calculate the y- components of tension from Newton's second law for equilibrium is,

$F_{\text{net},y}=F_y-mg$

• $F_{\text{net},y}$ is the net vertical force.
• $F_y$ is the vertical force on the strut from the wall
• $m$ is the mass of the strut
• g is the acceleration due to gravity

Since the strut is in equilibrium $F_{\text{net},y}=0$.

Formula to calculate the y- components of tension from Newton's second law for equilibrium is,

$\begin{array}{c}0=F_y-mg\\ F_y=mg\end{array}$

Weight of the strut acts downward and the force from the wall on the strut acts upward.

Conclusion:

The x and y-components of Newton's second law for equilibrium $F_x$ is $T$ and $F_y$ is $mg$.

(f)

To determine

The x- and y-components of the force on the hinge.

Answer to Problem 30P

:The x- and y-components of the force on the hinge $F_x$ is $132\text{N}$ and $F_y$ is $157\text{N}$

Explanation of Solution

Given info: The tension in the cable is $132\text{N}$., mass m is $16.0\text{kg}$, acceleration due to gravity g is $9.8\text{m}/{\text{s}}^{\text{2}}$.

Formula to calculate x component force exerted on the strut is,

$F_x=T$

Substitute $132\text{N}$ for $T$ in the above equation to find $F_x$.

$F_x=132\text{N}$

Formula to calculate y component force exerted on the strut is,

$F_y=mg$

Substitute $16.0\text{kg}$ for m and $9.8\text{m}/{\text{s}}^{\text{2}}$ for g in the above equation to find $F_y$

$\begin{array}{c}{F}_y=\left(16.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =157\text{N}\end{array}$

Conclusion:

The x- and y-components of the force on the hinge $F_x$ is $132\text{N}$ and $F_y$ is $157\text{N}$.

(g)

To determine

The advantages, if the cable is attached higher up on the wall.

Explanation of Solution

If the cable is inclined an angle $\theta$, the tension on it is directed both vertically and horizontally.

If the cable is attached higher up the wall, the vertical component of tension exists and thereby it can beam more weight.

Conclusion: Therefore, if the cable is attached higher up on the wall, the stress on the cable will be reduced and in turn the tension in the cable will be reduced.