Chapter 10, Problem 28P

A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 90.0 kg. The pivot under the left end exerts a normal force n₁ on the beam, and the second pivot located a distance ℓ = 4.00 m from the left end exerts a normal force n₂. A woman of mass m = 55.0 kg steps onto the left end of the beam and begins walking to the right as in Figure P10.28. The goal is to find the woman’s position when the beam begins to tip. (a) What is the appropriate analysis model for the beam before it begins to tip? (b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance x to the right of the first pivot, which is the origin. (c) Where is the woman when the normal force n₁ is the greatest? (d) What is n₁ when the beam is about to tip? (e) Use Equation 10.27 to find the value of n₂ when the beam is about to tip. (f) Using the result of part (d) and Equation 10.28, with torques computed around the second pivot, find the woman’s position x when the beam is about to tip. (g) Check the answer to part (e) by computing torques around the first pivot point.

Figure P10.28

To determine

The model of beam used for analysis before it begins to tip.

Answer to Problem 28P

The object is in static equilibrium before it begins to tip.

Explanation of Solution

Initially the beam is balanced. When an object is at rest the equilibrium maintained by the object is called static equilibrium.

Hence the object is in static equilibrium before it begins to tip.

Conclusion:

Therefore, the object is in static equilibrium before it begins to tip.

To determine

Sketch the force diagram for the beam.

Answer to Problem 28P

The force diagram of the beam is given below.

Explanation of Solution

Force diagram represents all the different kinds of forces acting on the given system. Weight of the women, $mg$ is experiencing at a distance $x$ from one end of the beam, and a force $Mg$ acting at a distance $3.00\text{m}$ from the same.

Conclusion:

The weight of the women is.

  $W_{\text{w}}=mg$        (I)

Here, $m$ is the mass of the women, $g$ is acceleration due to gravity.

The weight of the beam corresponding to center of mass is.

  $W_{\text{CM}}=Mg$        (II)

Substitute, $55.0\text{kg}$ for $m$ , $9.80\text{m}/{\text{s}}^2$ for $g$ in equation (I).

  $\begin{array}{c}{W}_{\text{w}}=\left(55.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\\ =539\text{N}\end{array}$

Substitute, $90.0\text{kg}$ for $m$ , $9.80\text{m}/{\text{s}}^2$ for $g$ in equation (II).

  $\begin{array}{c}{W}_{\text{CM}}=\left(90.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\\ =882\text{N}\end{array}$

The force diagram of the beam is given below.

To determine

The position of the women when the normal force is greatest.

Answer to Problem 28P

When the women is at $x=0$ the normal force $n_1$ is greatest.

Explanation of Solution

The normal force $n_1$ exert a clockwise torque about the right pivot, but all other forces are in counterclockwise direction, and the torque due to normal force $n_2$ is zero.

The normal force $n_1$ exerts maximum clockwise torque at  to keep the beam in rotational equilibrium. Thus, the normal force $n_1$ will be greatest when women is at $x=0$.

Conclusion:

Therefore, the normal force $n_1$ is greatest when the women is at $x=0$.

To determine

The value of $n_1$ when the beam is about to tip.

Answer to Problem 28P

The value of $n_1$ when the beam is about to tip zero.

Explanation of Solution

When the women walk from left end to right end she will reach at a point so that the beam start to rotate in clockwise direction about the right pivot. So the beam starts to lift up about the leftmost pivot, thus the normal force exerted by the pivot will be zero.

Thus the normal force $n_1$ will be zero when the beam is about to tip.

Conclusion:

Therefore, the value of $n_1$ when the beam is about to tip zero.

To determine

The value of $n_2$ when the beam is about to tip.

Answer to Problem 28P

The value of $n_2$ when the beam is about to tip is $\underset{\_}{1.42\times {10}^3\text{N}}$.

Explanation of Solution

Write the expression for the total force acting on $y$ direction.

  ${\displaystyle \sum F_y}=n_1+n_2-Mg-mg$        (III)

Since the force exerted on beam, and normal force $n_1$ will be zero when beam is about to tip. Substitute, $0$ for ${\displaystyle \sum F_y}$, and $n_1$ in equation (III).

  $\begin{array}{c}{\displaystyle \sum F_y}=0+n_2-Mg-mg\\ =0\end{array}$        (IV)

Rearrange equation (IV) to obtain an expression for $n_2$.

  $\begin{array}{c}{n}_2-Mg-mg=0\\ n_2=Mg+mg\end{array}$        (V)

Conclusion:

Substitute, $882\text{N}$ for ā$Mg$, $539\text{N}$ for $mg$ in equation (V) to find the normal force $n_2$.

  $\begin{array}{c}{n}_2=882\text{N}+539\text{N}\\ \text{=1}\text{.42}\times {\text{10}}^3\text{N}\end{array}$

Therefore, value of $n_2$ when the beam is about to tip is $\underset{\_}{1.42\times {10}^3\text{N}}$.

To determine

The position of women when the beam is about to tip.

Answer to Problem 28P

The position of women when the beam is about to tip is $\underset{\_}{5.64\text{m}}$.

Explanation of Solution

Write the expression for total torque in the beam.

  ${\displaystyle \sum \tau }=n_1{r}_1+n_2{r}_2+mg\left(4.00\text{m}-x\right)+\left(4.00\text{m}-\text{3}\text{.00}\text{m}\right)Mg$        (VI)

When the beam is about to the tip, the torque $n_1$, and the net torque about the pivot will be zero.

Substitute, $0$ for $n_1$, $r_1$, and $r_2$ in equation (VI) to obtain the value of $x$.

  $\begin{array}{c}\left(0\right)\left(0\right)+n_2\left(0\right)+mg\left(4.00\text{m}-x\right)+\left(4.00\text{m}-\text{3}\text{.00}\text{m}\right)Mg=0\\ mg\left(4.00\text{m}-x\right)+\left(4.00\text{m}-\text{3}\text{.00}\text{m}\right)Mg=0\\ mgx=\left(1.00\text{m}\right)Mg+\left(4.00\text{m}\right)mg\\ x=\left(1.00\text{m}\right)\frac{M}{m}+4.00\text{m}\end{array}$        (VII)

Conclusion:

Substitute, $55.0\text{kg}$ for $m$, $90.0\text{kg}$ for $M$ in equation (VII).

  $\begin{array}{c}x=\left(1.00\text{m}\right)\frac{90.0\text{kg}}{55.0\text{kg}}+4.00\text{m}\\ \text{=5}\text{.64}\text{m}\end{array}$

Therefore, the position of women when the beam is about to tip is $\underset{\_}{5.64\text{m}}$.

To determine

To check the answer in part (f) computing the torque around the first pivot point.

Answer to Problem 28P

The position of the women is $\underset{\_}{5.62\text{m}}$ similar to that of answer in part (f).

Explanation of Solution

Write the expression for the net torque about the left pivot.

  ${\displaystyle \sum \tau }=n_1{r}_1+n_2\left(4.00\text{m}\right)-Mg\left(3.00\text{m}\right)-mgx$        (VIII)

Conclusion:

The net torque about the left pivot is zero.

Substitute, $0$ for $n_1$, $r_1$, $882\text{N}$ for $Mg$, $539\text{N}$ for $mg$, $1.42\times {10}^3\text{N}$ for $n_2$ in equation (VIII).

$\begin{array}{c}0+\left(1.42\times {10}^3\text{N}\right)\left(4.00\text{m}\right)-\left(882\text{N}\right)\left(3.00\text{m}\right)-\left(539\text{N}\right)x=0\\ \left(882\text{N}\right)\left(3.00\text{m}\right)-\left(539\text{N}\right)x=-\left(1.42\times {10}^3\text{N}\right)\left(4.00\text{m}\right)\\ x=\frac{-3.03\times {10}^3\text{N}\cdot \text{m}}{-539\text{N}}\\ =5.62\text{m}\end{array}$

The position obtained in part (f) is $5.64\text{m}$, comparing both answers are almost same.

Therefore, the position of the women is $\underset{\_}{5.62\text{m}}$ similar to that of answer in part (f).