Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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.
### 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.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.