Question

Figure 12-84 shows a stationary arrangement of two crayon boxes and three cords. Box Ahas a mass of$\mathbf{11}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$ and is on a ramp at angle $\mathbf{\text{θ=}}\mathbf{30.}\mathbf{\text{0°}}$box Bhas a mass of$\mathbf{}\mathbf{7.00}\mathbf{\text{\hspace{0.17em}kg}}$ and hangs on a cord. The cord connected to box Ais parallel to the ramp, which is frictionless. (a) What is the tension in the upper cord, and (b) what angle does that cord make with the horizontal?

Short answer

1. Tension in the upper cord is$106\text{\hspace{0.17em}N}$.
2. Angle made by upper cord with the horizontal is $64.0°$.

Step-by-step solution

Understanding the given information

Mass of box A,${\text{m}}_{\text{A}}=11.0\text{\hspace{0.17em}kg}$

Mass of box B,${\text{m}}_{\text{B}}=7.00\text{\hspace{0.17em}kg}$

Angle made by ramp to the horizontal,$\mathrm{\theta }=30°$

Concept and formula used in the given question

Using the concept of equilibrium of forces, you can find the tension in the upper cord and the angle made by that cord with the horizontal. The equations used are given below.

$\begin{array}{l}\mathbf{Tcos}\mathbf{}\mathbf{\theta }\mathbf{=}{\mathbf{T}}_{\mathbf{A}}\mathbf{cos}\mathbf{}\mathbf{30}\mathbf{°}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{T}\mathbf{}\mathbf{sin\theta }\mathbf{=}{\mathbf{T}}_{\mathbf{A}}\mathbf{\times }\mathbf{sin}\mathbf{}\mathbf{(}\mathbf{30}\mathbf{°}\mathbf{)}\mathbf{}\mathbf{+}{\mathbf{T}}_{\mathbf{B}}\mathbf{}\end{array}$

(a) Calculation for the tension in the upper cord

We have,for the equilibrium of forces,

$\begin{array}{l}\mathrm{Tcos}{\mathrm{\theta }}^{\text{'}}={\mathrm{T}}_{\mathrm{A}}\cos 30°\\ \mathrm{T}{\mathrm{cos\theta }}^{\text{'}}={\mathrm{m}}_{\mathrm{A}}\times \mathrm{g}\times \sin (30°)\times \cos (30°)\left(1\right)\end{array}$

Also,

$\begin{array}{l}\mathrm{T}{\mathrm{sin\theta }}^{\text{'}}={\mathrm{T}}_{\mathrm{A}}\times \sin (30°)+{\mathrm{T}}_{\mathrm{B}}\\ \mathrm{T}{\mathrm{sin\theta }}^{\text{'}}={\mathrm{m}}_{\mathrm{A}}\times \mathrm{g}\times \sin (30°)\times \sin (30°)+{\mathrm{m}}_{\mathrm{B}}\times \mathrm{g}\left(2\right)\end{array}$

Now, solving eq. (1) and (2), for tension we get

$\text{T}=106\text{\hspace{0.17em}N}$

(b) Calculation for the angle does that cord make with the horizontal

Solving the above eq. (1) and eq. (2), for angle we get

${\mathrm{\theta }}^{\text{'}}=64.0°$