Chapter 4, Problem 9SP

A person stands on a scale, which then reads 600 N. (a) What force is exerted on the scale by the person? (b) What force is exerted on the person by the scale? (c) What would happen to the reading as the person began to jump straight up?

To determine

The magnitude of the force applied on the scale by the person if he stands on a scaleand the readingis $600\text{ N}$.

Answer to Problem 9SP

Solution:

$600\text{ N}$

Explanation of Solution

Given data:

The weight of the person is $600\text{ N}$.

The scale is at rest, so there will be no acceleration.

Formula used:

Write the expression for the first condition of force equilibrium:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or the horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or the vertical direction.

Explanation:

Draw the free body diagram of the scale when the person stands on it:

In the diagram, $W$ is the weight of the person and $F_N$ is the normal force exerted by the scale.

Recall the expression for the first condition of force equilibrium:

${\displaystyle \sum F_y}=0$

Consider that the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$F_N-W=0$

Substitute $600\text{ N }$ for $W$

$\begin{array}{c}{F}_N-600\text{ N}=0\\ F_N=600\text{ N}\end{array}$

From Newton’s third law, the force exerted on the scale by a person is equal and opposite to the weight of the person and it acts in the downward direction because the force applied by the person is in the downward direction. So, the magnitude of force applied by the person to the scale should be equal to the normal reaction force applied by the contact surface of the scale to the person.

Conclusion:

The magnitude of the force applied on the scale by the person is $600\text{ N}$.

To determine

The magnitude of the force applied on the person by the scale if he stands on a scale and the scale reads $600\text{ N}$.

Answer to Problem 9SP

Solution:

$600\text{ N}$

Explanation of Solution

Given data:

The weight of the person is $600\text{ N}$.

The scale is at rest, so there will be no acceleration.

Formula used:

Write the expression for the first condition of force equilibrium:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction or the horizontal direction and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction or the vertical direction.

Explanation:

Draw the free body diagram of the scale:

In the above diagram, $W$ is the weight of the person and $F_N$ is the normal force exerted by a scale.

Recall the expression for the first condition of force equilibrium:

${\displaystyle \sum F_y}=0$

Consider that the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$F_N-W=0$

Substitute $600\text{ N }$ for $W$

$\begin{array}{c}{F}_N-600\text{ N}=0\\ F_N=600\text{ N}\end{array}$

From Newton’sthird law, the force exerted on the person by the scale is equal and opposite to the weight of the person and it actsin the upward direction because the force exerted by the scale is in the upward direction.

Conclusion:

Therefore, the magnitude of the force applied on the person by the scale is $600\text{ N}$.

To determine

The reading on a scale when the person begins to jump straight up on a scale.

Answer to Problem 9SP

Solution:

Increase

Explanation of Solution

Given data:

The weight of the person is $600\text{ N}$.

The scale is at rest, so there will be no acceleration.

Formula used:

Write the expression for Newton’s second law of motion:

${\displaystyle \sum F}=ma$

Here, ${\displaystyle \sum F}$ is the sum of the forces, $m$ is the mass of the person, and $a$ is the acceleration in the upward direction.

Explanation:

Draw the free body diagram of the person when he begins to jump:

In the diagram given above, $W$ is the weight of the person, $F_N$ is the normal force exerted by the scale, $C.G_1$ is the center of gravity of the person when the person stands on a scale, $C.G_2$ is the center of gravity of the person when he begins to jump from the scale, $\Delta d$ is the displacement of the person or difference between $C.G_1$ and $C.G_2$, and $a$ is the upward acceleration of the person when the person begins to jump.

Recall the expression for Newton’s second law of motion along the vertical direction:

${\displaystyle \sum F}=ma$

Consider that the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$\begin{array}{c}{F}_N-W=ma\\ F_N=W+ma\end{array}$

Here, $ma$ is the additional force on the ground to accelerate the person in upward direction. Hence, the force exerted on the scale will increase the reading.

$$

Conclusion:

Therefore, as the person starts to jump straight up, the reading of the scale would increase.