Question

(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?
T w Both forces are equal.

(b) When the elevator is moving at a constant velocity upward, which is greater, T or w?
T w Both forces are equal.

(c) When the elevator is moving upward, but the acceleration is downward, which is greater, T or w?
T w Both forces are equal.

(d) Let the elevator have a mass of 1,600 kg and an upward acceleration of 2.5 m/s2. Find T.
N

Is your answer consistent with the answer to part (a)?
Yes No

(e) The elevator of part (d) now moves with constant upward velocity of 10 m/s. Find T.
N

Is your answer consistent with your answer to part (b)?
Yes No

(f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at 1.30 m/s2. Find T.
N

Is your answer consistent with your answer to part (c)?
Yes No

Answer 1

Concepts and reason

The concepts used to solve this question is free body diagram and Newton’s laws of motion.

Initially, draw the free body diagram and then use Newton’s laws of motion to determine the tension in the cable.

Fundamentals

”The motion is defined as the change in position of the object from one point to another. The object changes its position because there is a net force acting on the object”

The magnitude of the net force is proportional to the mass of the object and acceleration of the object.

The magnitude of the net force acting on the object is,

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

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

(a)

The free body diagram for the elevator is drawn below.

From, the above diagram, the net force acting on the elevator is,

$T = mg + ma$ …… (1)

Here, $m$ is the mass of the elevator, $g$ is the acceleration due to gravity and $a$ is the acceleration of the elevator.

The weight of the elevator is,

$W = mg$

Substitute $W$ for $mg$ in the equation (1).

$T = W + ma$

Thus, from the above result, the tension in the cable is greater

(b)

At constant velocity, the acceleration of the elevator is zero.

Therefore,

$a = 0$

The tension in the cable when the elevator is moving with constant velocity is,

$T = W + ma$ …… (2)

Substitute $0{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the equation (2).

$\begin{array}{c}\\T = W + ma\\\\ = W + 0\\\\ = W\\\end{array}$

(c)

The free body diagram of the elevator is drawn below.

The net force acting on the elevators is,

$\begin{array}{c}\\mg - T = ma\\\\T = mg - ma\\\\ = W - ma\\\end{array}$

Hence, from the above result, the tension in the cable is less than the weight of the elevator.

(d.1)

The free body diagram of the elevator is drawn below.

The net force acting on the elevators is,

$T = mg + ma$

Substitute $1600{\rm{ kg}}$ for $m$ , $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ and $2.5{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the above equation.

$\begin{array}{c}\\T = mg + ma\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right) + \left( {1600{\rm{ kg}}} \right)\left( {2.5{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 19696{\rm{ N}}\\\end{array}$

(d.2)

The weight of the elevator is,

$W = mg$ …… (3)

Substitute $1600{\rm{ kg}}$ for $m$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation (3).

$\begin{array}{c}\\W = mg\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 15696{\rm{ N}}\\\end{array}$

Hence, from the above two result

$T > W$

Hence, the answer is consistent with part (a).

(e.1)

At constant velocity, the acceleration of the elevator is zero.

Therefore,

$a = 0$

The tension in the cable when the elevator is moving with constant velocity is,

$T = W + ma$ …… (4)

Substitute $0{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the equation (2).

$\begin{array}{c}\\T = W + ma\\\\ = W + 0\\\\ = W\\\end{array}$

The above equation is modified as,

$T = mg$ …… (5)

Substitute $1600{\rm{ kg}}$ for $m$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation (5).

$\begin{array}{c}\\T = mg\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 15696{\rm{ N}}\\\end{array}$

(e.2)

The weight of the elevator is,

$W = mg$

Substitute $1600{\rm{ kg}}$ for $m$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the above equation.

$\begin{array}{c}\\W = mg\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 15696{\rm{ N}}\\\end{array}$

Hence, from the above two result

$T = W$

Hence, the answer is consistent with part (b).

(f.1)

The free body diagram of elevator is drawn below.

The net force acting on the elevators is,

$T = mg - ma$ …… (6)

Substitute $1600{\rm{ kg}}$ for $m$ , $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ and $2.5{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the above equation.

$\begin{array}{c}\\T = mg - ma\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right) - \left( {1600{\rm{ kg}}} \right)\left( {2.5{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 11696{\rm{ N}}\\\end{array}$

(f.2)

The weight of the elevator is,

$W = mg$ …… (7)

Substitute $1600{\rm{ kg}}$ for $m$ and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation (7).

$\begin{array}{c}\\W = mg\\\\ = \left( {1600{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 15696{\rm{ N}}\\\end{array}$

Hence, from the above two result

$T < W$

Hence, the answer is consistent with part (c).

Ans: Part a

The tension in the cable is greater than the weight of the elevator.