Chapter 4, Problem 35P

(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater. T or w? (b) When the elevator is moving at a constant velocity upward, which is greater. T or w² (c) When the elevator is moving upward, but the acceleration is downward, which is greater. T or w? (d) Let the elevator have a mass of 1500 kg and an upward acceleration of 2.5 m/s². Find T. Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of 10 m/s. Find T. Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at 1.50 m/s². Find T Is your answer consistent with your answer to part (c)?

To determine

To compare the magnitude of tension and weight of the elevator accelerating upward.

Answer to Problem 35P

Solution:

The magnitude of tension is greater than the weight of the elevator.

Explanation of Solution

Given Info:

The mass of the elevator is m.

Write the formula to calculate weight of the elevator.

$W=mg$ (I)

• W is the weight of the elevator
• m is the mass of the elevator
• g is the acceleration due to gravity

  Write the equation of motion of the elevator.

  $T-mg=ma$

• T is the tension on the elevator
• a is the acceleration of the elevator

  Rewrite the equation of motion in terms of the tension.

  $T=mg+ma$ (II)

  Compare (I) and (II).

  Thus the tension is greater than the weight of the elevator.

  Conclusion:

  Therefore, the magnitude of tension is greater than the weight of the elevator.

To determine

To determine: To compare the magnitude of the tension and weight when the elevator is moving with constant velocity.

Answer to Problem 35P

Solution:

The magnitude of tension and weight of the elevator is same.

Explanation of Solution

Explanation:

Given Info:

When the elevator is moving with constant velocity in upward direction then the acceleration of the elevator would become zero.

Write the formula to calculate the tension on the elevator.

$T=mg$ (III)

Compare (I) and (III)

Thus the elevator whose upward velocity is constant would have same magnitude for tension and weight.

Conclusion:

Therefore, the magnitude of tension and weight of the elevator is same.

To determine

To determine: To compare the magnitude of the tension and weight while elevator is moving upward but acceleration is downward.

Answer to Problem 35P

Answer:

The weight of the elevator would have greater magnitude than the tension.

Explanation of Solution

Explanation:

When elevator is moving upward the weight of the elevator always points towards downward. The weight of any object or body arises because of the gravitational force acting on the body or object by the earth.

In this case elevator is moving upward but acceleration is downward. This downward acceleration would added to the acceleration due to gravity and net acceleration increases an consequently weight also increases. Thus the weight of the elevator would have greater magnitude than the tension.

Conclusion:

Therefore, the weight of the elevator would have greater magnitude than the tension.

To determine

To determine: To determine the magnitude of tension and compare the answer with (a).

Answer to Problem 35P

Solution:

The magnitude of tension is greater than the weight of the elevator which is consistent with (a) and the magnitude of tension is $\text{1}\text{.85}\times {\text{10}}^4\text{N}$.

Explanation of Solution

Explanation:

Given Info:

The mass of the elevator is $1500\text{kg}$ , the acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$ and upward acceleration is $2.5\text{m}/{\text{s}}^{\text{2}}$.

Write the formula to calculate weight of the elevator.

$W=mg$

Substitute $1500\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate W.

$\begin{array}{c}W=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N}\end{array}$

Write the formula to calculate tension on the elevator.

$T=mg+ma$

Substitute $1500\text{kg}$ for m, $2.5\text{m}/{\text{s}}^{\text{2}}$ for a and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate T.

$\begin{array}{c}T=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)+\left(1500\text{kg}\right)\left(2.5\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N+3750}\text{N}\\ \text{=18450}\text{N}\\ \text{=1}\text{.85}\times {\text{10}}^4\text{N}\end{array}$

Thus the tension is greater than the weight of the elevator.

Conclusion:

Therefore, the magnitude of tension is greater than the weight of the elevator which is consistent with (a) and the magnitude of tension is $\text{1}\text{.85}\times {\text{10}}^4\text{N}$.

To determine

To determine: To compare the magnitude of the tension and weight when the elevator is moving with constant velocity and compare with (b).

Answer to Problem 35P

Solution:

The magnitude of tension and weight of the elevator is same which is consistent with (b) and the value is $\text{1}\text{.47}\times {\text{10}}^4\text{N}$.

Explanation of Solution

Explanation:

Given Info:

When the elevator is moving with constant velocity in upward direction then the acceleration of the elevator would become zero.

Write the formula to calculate weight of the elevator.

$W=mg$

Substitute $1500\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate W.

$\begin{array}{c}W=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N}\end{array}$

Write the formula to calculate tension on the elevator.

$T=mg+ma$

Substitute $1500\text{kg}$ for m, $2.5\text{m}/{\text{s}}^{\text{2}}$ for a and $0\text{m}/{\text{s}}^{\text{2}}$ for g to calculate T.

$\begin{array}{c}T=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)+\left(1500\text{kg}\right)\left(0\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N+0}\text{N}\\ \text{=14700}\text{N}\\ \text{=1}\text{.47}\times {\text{10}}^4\text{N}\end{array}$

Thus the elevator whose upward velocity is constant would have same magnitude for tension which is consistent with (b) and weight.

Conclusion:

Therefore, the magnitude of tension and weight of the elevator is same and the value is $\text{1}\text{.47}\times {\text{10}}^4\text{N}$.

To determine

To determine: To determine the magnitude of tension and compare the answer with (c).

Answer to Problem 35P

Solution:

The magnitude of tension is less than the weight of the elevator which is consistent with (c) and the magnitude of tension is $\text{1}\text{.25}\times {\text{10}}^4\text{N}$.

Explanation of Solution

Explanation:

Given Info:

The mass of the elevator is $1500\text{kg}$ , the acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$ and downward acceleration is $1.50\text{m}/{\text{s}}^{\text{2}}$.

Since the acceleration is downward the magnitude taken as negative.

Write the formula to calculate weight of the elevator.

$W=mg$

Substitute $1500\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate W.

$\begin{array}{c}W=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N}\end{array}$

Write the formula to calculate tension on the elevator.

$T=mg+ma$

Substitute $1500\text{kg}$ for m, $-1.50\text{m}/{\text{s}}^{\text{2}}$ for a and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate T.

$\begin{array}{c}T=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)+\left(1500\text{kg}\right)\left(-1.50\text{m}/{\text{s}}^{\text{2}}\right)\\ =14700\text{N-2250}\text{N}\\ \text{=12450}\text{N}\\ \text{=1}\text{.25}\times {\text{10}}^4\text{N}\end{array}$

Thus the tension is less than the weight of the elevator.

Conclusion:

Therefore, the magnitude of tension is less than the weight of the elevator which is consistent with (c) and the magnitude of tension is $\text{1}\text{.25}\times {\text{10}}^4\text{N}$