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Sagot :
Let's address question 30 first:
Question 30: An 80 kg painter climbs 85% of the way up a uniform ladder of 25 kg before it starts to slip backwards along the ground. What is the coefficient of static friction between the ground and ladder?
### Step-by-Step Solution
1. Mass of Painter and Ladder:
- The mass of the painter is [tex]\(80 \text{ kg}\)[/tex].
- The mass of the ladder is [tex]\(25 \text{ kg}\)[/tex].
2. Ladder Length:
- For simplicity, let's assume the ladder's total length is [tex]\(1 \text{ m}\)[/tex].
3. Distance Climbed:
- The distance climbed by the painter up the ladder is [tex]\(85\%\)[/tex] of the ladder's length.
- Therefore, the distance climbed is [tex]\(0.85 \text{ m}\)[/tex].
4. Calculating Forces Due to Gravity (Weights):
- The force due to gravity for the painter (weight) is calculated as:
[tex]\( \text{Weight of Painter} = 80 \text{ kg} \times 9.8 \text{ m/s}^2 = 784 \text{ N}\)[/tex].
- The force due to gravity for the ladder (weight) is calculated as:
[tex]\( \text{Weight of Ladder} = 25 \text{ kg} \times 9.8 \text{ m/s}^2 = 245 \text{ N}\)[/tex].
5. Center of Mass of the Ladder:
- Since the ladder is uniform, its center of mass is at its midpoint.
- Therefore, the center of mass is at [tex]\(0.5 \text{ m}\)[/tex].
6. Torque Calculation About the Base of the Ladder:
- The torque due to the painter is calculated as:
[tex]\( \text{Torque by Painter} = 784 \text{ N} \times 0.85 \text{ m} = 666.4 \text{ Nm}\)[/tex].
- The torque due to the ladder is calculated as:
[tex]\( \text{Torque by Ladder} = 245 \text{ N} \times 0.5 \text{ m} = 122.5 \text{ Nm}\)[/tex].
7. Total Torque:
- The total torque trying to rotate the ladder is:
[tex]\( \text{Total Torque} = 666.4 \text{ Nm} + 122.5 \text{ Nm} = 788.9 \text{ Nm}\)[/tex].
8. Normal Force:
- The painter and the ladder exert a normal force on the ground.
- This normal force is the sum of their weights divided by 2 (assuming equilibrium condition and uniform weight distribution):
[tex]\( \text{Normal Force} = \frac{784 \text{ N} + 245 \text{ N}}{2} = 514.5 \text{ N}\)[/tex].
9. Coefficient of Static Friction Calculation:
- The coefficient of static friction is given by the ratio of the total torque to the normal force:
[tex]\[ \text{Coefficient of Static Friction} = \frac{\text{Total Torque}}{\text{Normal Force}} = \frac{788.9 \text{ Nm}}{514.5 \text{ N}} \approx 1.533.\][/tex]
Thus, the coefficient of static friction between the ground and the ladder is approximately [tex]\(1.533\)[/tex].
---
Now moving on to Question 31 which seems to be disconnected from question 30 and requires additional information for a complete solution. Let's start with the given details:
Question 31: A car of mass 1500 kg accelerates from rest.
Since the question appears incomplete, we can only infer some scenarios. Let's consider a common question type—calculating the force needed if the car accelerates at a given rate.
### Complete the information:
If a car of mass [tex]\(1500 \text{ kg}\)[/tex] accelerates with an acceleration [tex]\(a\)[/tex]:
### Formula:
The force [tex]\(F\)[/tex] exerted can be found using Newton's second law:
[tex]\[ F = ma \][/tex]
### Example Calculation:
If we assume [tex]\(a = 2 \text{ m/s}^2\)[/tex] (if not provided, replace [tex]\(2 \text{ m/s}^2\)[/tex] with the actual acceleration value):
[tex]\[ F = 1500 \text{ kg} \times 2 \text{ m/s}^2 = 3000 \text{ N} \][/tex]
Thus, the force required for the car to accelerate at [tex]\(2 \text{ m/s}^2\)[/tex] is [tex]\(3000 \text{ N}\)[/tex].
Question 30: An 80 kg painter climbs 85% of the way up a uniform ladder of 25 kg before it starts to slip backwards along the ground. What is the coefficient of static friction between the ground and ladder?
### Step-by-Step Solution
1. Mass of Painter and Ladder:
- The mass of the painter is [tex]\(80 \text{ kg}\)[/tex].
- The mass of the ladder is [tex]\(25 \text{ kg}\)[/tex].
2. Ladder Length:
- For simplicity, let's assume the ladder's total length is [tex]\(1 \text{ m}\)[/tex].
3. Distance Climbed:
- The distance climbed by the painter up the ladder is [tex]\(85\%\)[/tex] of the ladder's length.
- Therefore, the distance climbed is [tex]\(0.85 \text{ m}\)[/tex].
4. Calculating Forces Due to Gravity (Weights):
- The force due to gravity for the painter (weight) is calculated as:
[tex]\( \text{Weight of Painter} = 80 \text{ kg} \times 9.8 \text{ m/s}^2 = 784 \text{ N}\)[/tex].
- The force due to gravity for the ladder (weight) is calculated as:
[tex]\( \text{Weight of Ladder} = 25 \text{ kg} \times 9.8 \text{ m/s}^2 = 245 \text{ N}\)[/tex].
5. Center of Mass of the Ladder:
- Since the ladder is uniform, its center of mass is at its midpoint.
- Therefore, the center of mass is at [tex]\(0.5 \text{ m}\)[/tex].
6. Torque Calculation About the Base of the Ladder:
- The torque due to the painter is calculated as:
[tex]\( \text{Torque by Painter} = 784 \text{ N} \times 0.85 \text{ m} = 666.4 \text{ Nm}\)[/tex].
- The torque due to the ladder is calculated as:
[tex]\( \text{Torque by Ladder} = 245 \text{ N} \times 0.5 \text{ m} = 122.5 \text{ Nm}\)[/tex].
7. Total Torque:
- The total torque trying to rotate the ladder is:
[tex]\( \text{Total Torque} = 666.4 \text{ Nm} + 122.5 \text{ Nm} = 788.9 \text{ Nm}\)[/tex].
8. Normal Force:
- The painter and the ladder exert a normal force on the ground.
- This normal force is the sum of their weights divided by 2 (assuming equilibrium condition and uniform weight distribution):
[tex]\( \text{Normal Force} = \frac{784 \text{ N} + 245 \text{ N}}{2} = 514.5 \text{ N}\)[/tex].
9. Coefficient of Static Friction Calculation:
- The coefficient of static friction is given by the ratio of the total torque to the normal force:
[tex]\[ \text{Coefficient of Static Friction} = \frac{\text{Total Torque}}{\text{Normal Force}} = \frac{788.9 \text{ Nm}}{514.5 \text{ N}} \approx 1.533.\][/tex]
Thus, the coefficient of static friction between the ground and the ladder is approximately [tex]\(1.533\)[/tex].
---
Now moving on to Question 31 which seems to be disconnected from question 30 and requires additional information for a complete solution. Let's start with the given details:
Question 31: A car of mass 1500 kg accelerates from rest.
Since the question appears incomplete, we can only infer some scenarios. Let's consider a common question type—calculating the force needed if the car accelerates at a given rate.
### Complete the information:
If a car of mass [tex]\(1500 \text{ kg}\)[/tex] accelerates with an acceleration [tex]\(a\)[/tex]:
### Formula:
The force [tex]\(F\)[/tex] exerted can be found using Newton's second law:
[tex]\[ F = ma \][/tex]
### Example Calculation:
If we assume [tex]\(a = 2 \text{ m/s}^2\)[/tex] (if not provided, replace [tex]\(2 \text{ m/s}^2\)[/tex] with the actual acceleration value):
[tex]\[ F = 1500 \text{ kg} \times 2 \text{ m/s}^2 = 3000 \text{ N} \][/tex]
Thus, the force required for the car to accelerate at [tex]\(2 \text{ m/s}^2\)[/tex] is [tex]\(3000 \text{ N}\)[/tex].
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