Certainly! Let's analyze the problem step-by-step, incorporating all the necessary physics principles and geometry.
### Problem Breakdown:
You have a sphere:
- Weight [tex]\( W = 40 \)[/tex] N
- Radius [tex]\( r = 30 \)[/tex] cm
This sphere is in contact with a smooth vertical wall and is supported by a string:
- Length of the string [tex]\( L = 20 \)[/tex] cm
To solve this problem, we'll need to find:
1. The tension in the string.
2. The horizontal reaction force from the wall.
### Step-by-step Solution:
#### Step 1: Convert Units
First, convert all given lengths to meters to maintain consistency in the units.
- Radius: [tex]\( r = 30 \)[/tex] cm = [tex]\( 0.30 \)[/tex] m
- String length: [tex]\( L = 20 \)[/tex] cm = [tex]\( 0.20 \)[/tex] m
#### Step 2: Geometry of the System
Visualize the setup:
- The string forms a right-angled triangle with the vertical wall and the radius of the sphere.
- The hypotenuse of this triangle is the sum of the radius of the sphere and the length of the string.
Therefore, the hypotenuse [tex]\( h \)[/tex] is:
[tex]\[ h = r + L = 0.30 \text{ m} + 0.20 \text{ m} = 0.50 \text{ m} \][/tex]
The adjacent side to the angle [tex]\( \theta \)[/tex] is the radius of the sphere:
[tex]\[ \text{adjacent} = r = 0.30 \text{ m} \][/tex]
#### Step 3: Determine the Angle [tex]\( \theta \)[/tex]
Using the cosine function in the right triangle:
[tex]\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{0.30 \text{ m}}{0.50 \text{ m}} = 0.6 \][/tex]
Therefore,
[tex]\[ \theta = \cos^{-1}(0.6) \][/tex]
#### Step 4: Tension in the String
To find the tension in the string, consider the vertical forces:
- The vertical component of the tension [tex]\( T \cos(\theta) \)[/tex] must balance the weight of the sphere [tex]\( W \)[/tex].
- Hence, [tex]\( T \cos(\theta) = W \)[/tex].
Solving for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{W}{\cos(\theta)} = \frac{40 \text{ N}}{0.6} = 66.67 \text{ N} \][/tex]
#### Step 5: Reaction Due to the Wall
The horizontal component of the tension balances the horizontal reaction from the wall.
- The horizontal reaction force [tex]\( F_{\text{horizontal}} \)[/tex] satisfies:
[tex]\[ F_{\text{horizontal}} = T \sin(\theta) \][/tex]
First, find [tex]\( \sin(\theta) \)[/tex] using [tex]\( \cos(\theta) \)[/tex]:
[tex]\[ \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - (0.6)^2} = \sqrt{0.64} = 0.8 \][/tex]
Therefore, the horizontal reaction force is:
[tex]\[ F_{\text{horizontal}} = T \sin(\theta) = 66.67 \text{ N} \times 0.8 = 53.33 \text{ N} \][/tex]
### Final Answers:
(a) The tension in the string is approximately [tex]\( 66.67 \)[/tex] N.
(b) The horizontal reaction force from the wall is approximately [tex]\( 53.33 \)[/tex] N.
