Question

Question: In Fig. 12-26, a uniform sphere of mass m = 0.85 m and radius r = 4.2 mis held in place by a massless rope attached to a frictionless wall a distance L = 8.0 cm above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.

Short answer

Answer:

1. The tension in the rope is 9.4 N
2. The force on the sphere from the wall is 4.4 N

Step-by-step solution

Understanding the given information 

The mass of the sphere,$m=0.85\text{kg}.$

The radius of the sphere,$r=4.2\text{cm}=0.042\text{m}.$

The distance between the center of the sphere and the point on the wall, L = 8.0 cm

Concept and formula used in the given question

Using the free body diagram and Newton’s second law and condition for equilibrium, you will write two force equations for vertical and horizontal direction. Using one of these equations, you can find the tension in the rope and using the second equation, you can find the force on the sphere by the wall. The law used is stated below.

Newton’s second law:

$F_{net}=ma$

(a) Calculation for the tension in the rope

The free body diagram:

Applying Newton’s second law

$$\begin{gathered} T\text{cos}\theta -mg=0······\left(1\right) \\ {F}_N-T\text{sin}\theta =0······\left(2\right) \end{gathered}$$

From equation (1),

$T=\frac{mg}{\text{cos}\theta }$

But from the figure,

$\text{cos}\theta =\frac{L}{\sqrt{L^2+r^2}}$

So,

$$\begin{gathered} T=\frac{mg}{\frac{L}{\sqrt{L^2+r^2}}} \\ =\frac{\left(0.85\right)\left(9.8\right)\sqrt{{0.08}^2+{0.042}^2}}{0.08} \\ =9.4\text{N} \end{gathered}$$

Hence, the tension in the rope is 9.4 N

(b) Calculation for the force on the sphere from the wall  

From the second equation, the normal force is,

$F_N=T\text{sin}\theta$

But from the figure,

$\text{sin}\theta =\frac{r}{\sqrt{L^2+r^2}}$

So,

$$\begin{gathered} F_N=T\frac{r}{\sqrt{L^2+r^2}} \\ =\frac{9.4\left(0.042\right)}{\sqrt{{0.08}^2+{0.042}^2}} \\ =4.369 \\ ~4.4\text{N} \end{gathered}$$

Hence, the force on the sphere from the wall is 4.4N