Question

In Figure 5-44, elevator cabs Aand Bare connected by a short cable and can be pulled upward or lowered by the cable above cab A. Cab Ahas mass 1700 kg; cab Bhas mass 1300 kg. A 12.0 kgbox of catnip lies on the floor of cab A. The tension in the cable connecting the cabs is$\mathbf{1}\mathbf{.}\mathbf{91}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{N}$⁴. What is the magnitude of the normal force on the box from the floor?

Short answer

The magnitude of the normal force acting on the box from the floor of cab A is 176N.

Step-by-step solution

Given information

• The masses of the elevator cabs are,$M_{\mathit{A}}=1700\mathrm{kg},{\mathrm{M}}_{\mathrm{B}}=1300\mathrm{kg}$
• The tension in the cable connecting the cabs is,$T=1.91\times {10}^4\mathrm{N}$
• Mass of the box,$M=12.0\mathrm{kg}$

Understanding the concept of Newton’s second law

Newton’s second law states that the net force acting on the body is equal to the vector sum of all the forces. The force is also equal to the product of mass and acceleration of the object.

Using the free body diagram and Newton’s second law, we can find the magnitude of the normal force acting on the box from the floor of cab A.

Formulae:

Newton’s second law is,

$F_{net}=\sum _{}Ma$

Calculating the acceleration of the cab

The forces acting on the cab B are Tension in the upward direction and weight in the downward direction. The net acceleration is in the upward direction. We can draw FBD for cab B to understand this situation and find the acceleration.

From FBD, we can write as,

$$\begin{gathered} {\mathrm{M}}_{\mathrm{B}}\mathrm{a}=\mathrm{T}-{\mathrm{M}}_{\mathrm{B}}\mathrm{g} \\ \mathrm{a}=\frac{\mathrm{T}-{\mathrm{M}}_{\mathrm{B}}\mathrm{g}}{{\mathrm{M}}_{\mathrm{B}}} \\ =\frac{\left[\left(1.91\times {10}^4\mathrm{N}\right)-\left(1300\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)\right]}{1300\mathrm{kg}} \\ =4.89\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the cab is $4.89\mathrm{m}/{\mathrm{s}}^2$. Both cab A and B will have the same acceleration.

Calculating the normal force on the box

Since, both the cabs have the same acceleration, the box in cab A will also have the same acceleration. The forces on the box are, the weight of the box in the downward direction, and the normal force acting on the box. Free body diagram for box placed on the elevator cab A:

From FBD, we can write as,

$$\begin{gathered} F_N-Mg=Ma \\ {F}_N=M\left(g+a\right) \\ =12\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2+4.89\mathrm{m}/{\mathrm{s}}^2\right) \\ =176\mathrm{N} \end{gathered}$$

Therefore, the normal force on the box is 176 N.