Chapter 4, Problem 6SP

A 85-kg man is in an elevator that is accelerating downward at the rate of 1.3 m/s².

1. a. What is the true weight of the man in newtons?
2. b. What is the net force acting on the man required to produce the acceleration?
3. c. What is the force exerted on theman’s feet by the floor of the elevator?
4. d. What is the apparent weight of the man in newtons? (This is the weight that would be read on the scale dial if the man were standing on a bathroom scale in the accelerating elevator.)
5. e. How would your answers to parts b through d change if the elevator were accelerating upward with an acceleration of 1.3 m/s²?

To determine

The true weight of the man of mass $85\text{kg}$ in newtons.

Answer to Problem 6SP

The true weight of the man of mass $85\text{kg}$ in newtons is $833\text{N}$.

Explanation of Solution

Given info: The mass of the man is $85\text{kg}$.

Write the expression for the weight.

$W=mg$

Here,

$W$ is the weight of the man

$m$ is the mass of the man

$g$ is the acceleration of the man

Substitute $85\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to get $W$.

$\begin{array}{c}W=\left(85\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =833\text{N}\end{array}$

Conclusion:

Thus, the true weight of the man of mass $85\text{kg}$ in newtons is $833\text{N}$.

To determine

The net force acting on the man required to produce the acceleration.

Answer to Problem 6SP

The net force acting on the man required to produce the acceleration is $110.5\text{N}$.

Explanation of Solution

Given info: The mass of the man is $85\text{kg}$ and acceleration of the man is $1.3\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the net force on the man.

$F_{\text{net}}=ma$

Here,

$F_{\text{net}}$ is the net force acting on the man

$a$ is the acceleration of the man

Substitute $85\text{kg}$ for $m$ and $1.3\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=\left(85\text{kg}\right)\left(1.3\text{m}/{\text{s}}^{\text{2}}\right)\\ =110.5\text{N}\end{array}$

Since the acceleration is in the downward direction , the net force will also in the downward direction.

Conclusion:

Thus, the net force acting on the man required to produce the acceleration is $110.5\text{N}$.

To determine

The fore exerted on the man’s feet by the floor of the elevator.

Answer to Problem 6SP

The force exerted on the man’s feet by the floor of the elevator is $722.5\text{N}$.

Explanation of Solution

Force exerted on the man’s feet by the floor is in the upward direction.

Write the expression for the net force acting on the man

$F_{\text{net}}=F_g-F_N$

Here,

$F_N$ is the normal force acting on the mass by the floor of the elevator

The negative sign in the above equation indicate that the normal force is opposite to gravitational force. Since gravitational force is in downward direction normal force is in the upward direction.

Substitute $110.5\text{N}$ for $F_{\text{net}}$ and $833\text{N}$ for $F_g$ in the above equation to get $F_N$

$\begin{array}{c}110.5\text{N}=833\text{N}-F_N\\ F_N=722.5\text{N}\end{array}$

Conclusion:

Thus, the force exerted on the man’s feet by the floor of the elevator is $722.5\text{N}$.

To determine

The apparent weight of the man in newtons.

Answer to Problem 6SP

The apparent weight of the man in newtons is $722.5\text{N}$.

Explanation of Solution

The apparent weight is the weight is the weight experienced by the man in the elevator.

This is weight measured by a weighing apparatus in an elevator. According to newtons third law

This should be equal to normal force acting on the man by the floor.

The normal force on the man by the floor is equal to $722.5\text{N}$, Therefore apparent weight of the man is equal to $722.5\text{N}$.

Conclusion:

Thus, the apparent weight of the man in newtons is $722.5\text{N}$.

To determine

The apparent weight of the man if the elevator is accelerating upward with an acceleration of $1.3\text{m}/{\text{s}}^{\text{2}}$.

Answer to Problem 6SP

The apparent weight of the man if the elevator is accelerating upward with an acceleration of $1.3\text{m}/{\text{s}}^{\text{2}}$ is $943.5\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression for the normal force acting on the man.

$F_N=F_g+F_{\text{net}}$

Substitute $110.5\text{N}$ for $F_{\text{net}}$ and $833\text{N}$ for $F_g$ in the above equation to get $F_N$.

$\begin{array}{c}{F}_N=833\text{N}+110.5\text{N}\\ \text{=943}\text{.5}\text{N}\end{array}$

The apparent weight is the weight is the weight experienced by the man in the elevator.

This is weight measured by a weighing apparatus in an elevator. According to newtons third law This should be equal to normal force acting on the man by the floor.

The normal force on the man by the floor is equal to $\text{943}\text{.5}\text{N}$, Therefore apparent weight of the man is equal to $\text{943}\text{.5}\text{N}$.

Conclusion:

Thus, the apparent weight of the man in newtons is $\text{943}\text{.5}\text{N}$.