Answer :
Sure! Let's solve these numerical problems step by step.
### Problem (a):
An effort of 200 N is used to lift a load of 800 N by using a lever. The load is at a distance of 40 cm from the fulcrum. Find the effort distance.
Solution:
1. We know the effort (E) is 200 N.
2. The load (L) is 800 N.
3. The load distance (dL) is 40 cm. Converting it to meters, we get 0.4 meters.
Using the principle of moments (also known as the lever principle):
[tex]\[ \text{Moment} = \text{Force} \times \text{Distance} \][/tex]
For the system to be in equilibrium:
[tex]\[ \text{Effort} \times \text{Effort Distance} = \text{Load} \times \text{Load Distance} \][/tex]
Let's denote the effort distance by [tex]\( dE \)[/tex]:
[tex]\[ 200 \times dE = 800 \times 0.4 \][/tex]
Now, solve for [tex]\( dE \)[/tex]:
[tex]\[ dE = \frac{800 \times 0.4}{200} \][/tex]
[tex]\[ dE = \frac{320}{200} \][/tex]
[tex]\[ dE = 1.6 \text{ meters} \][/tex]
Converting meters to centimeters:
[tex]\[ dE = 1.6 \times 100 = 160 \text{ cm} \][/tex]
Answer: The effort distance is 160 cm.
### Problem (b):
Ram with a weight of 400 N and Shyam with a weight of 500 N are willing to play on a see-saw. If Ram is at a distance of 2 meters from the fulcrum, how far from the fulcrum should Shyam sit to balance Ram?
Solution:
1. Ram's weight (WR) is 400 N.
2. Ram's distance from the fulcrum (dR) is 2 meters.
3. Shyam's weight (WS) is 500 N.
For the see-saw to be in balance:
[tex]\[ \text{Moment of Ram} = \text{Moment of Shyam} \][/tex]
[tex]\[ WR \times dR = WS \times dS \][/tex]
Where [tex]\( dS \)[/tex] is the distance Shyam should sit from the fulcrum.
[tex]\[ 400 \times 2 = 500 \times dS \][/tex]
Now solve for [tex]\( dS \)[/tex]:
[tex]\[ dS = \frac{400 \times 2}{500} \][/tex]
[tex]\[ dS = \frac{800}{500} \][/tex]
[tex]\[ dS = 1.6 \text{ meters} \][/tex]
Answer: Shyam should sit 1.6 meters from the fulcrum to balance Ram.
### Problem (c):
The mechanical advantage of a simple machine is 3 and its velocity ratio is 4. Find its efficiency.
Solution:
1. Mechanical Advantage (MA) is 3.
2. Velocity Ratio (VR) is 4.
The efficiency of a simple machine is given by:
[tex]\[ \text{Efficiency} (\%) = \left( \frac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} \right) \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = \left( \frac{3}{4} \right) \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = 0.75 \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = 75 \][/tex]
Answer: The efficiency of the machine is 75%.
### Problem (a):
An effort of 200 N is used to lift a load of 800 N by using a lever. The load is at a distance of 40 cm from the fulcrum. Find the effort distance.
Solution:
1. We know the effort (E) is 200 N.
2. The load (L) is 800 N.
3. The load distance (dL) is 40 cm. Converting it to meters, we get 0.4 meters.
Using the principle of moments (also known as the lever principle):
[tex]\[ \text{Moment} = \text{Force} \times \text{Distance} \][/tex]
For the system to be in equilibrium:
[tex]\[ \text{Effort} \times \text{Effort Distance} = \text{Load} \times \text{Load Distance} \][/tex]
Let's denote the effort distance by [tex]\( dE \)[/tex]:
[tex]\[ 200 \times dE = 800 \times 0.4 \][/tex]
Now, solve for [tex]\( dE \)[/tex]:
[tex]\[ dE = \frac{800 \times 0.4}{200} \][/tex]
[tex]\[ dE = \frac{320}{200} \][/tex]
[tex]\[ dE = 1.6 \text{ meters} \][/tex]
Converting meters to centimeters:
[tex]\[ dE = 1.6 \times 100 = 160 \text{ cm} \][/tex]
Answer: The effort distance is 160 cm.
### Problem (b):
Ram with a weight of 400 N and Shyam with a weight of 500 N are willing to play on a see-saw. If Ram is at a distance of 2 meters from the fulcrum, how far from the fulcrum should Shyam sit to balance Ram?
Solution:
1. Ram's weight (WR) is 400 N.
2. Ram's distance from the fulcrum (dR) is 2 meters.
3. Shyam's weight (WS) is 500 N.
For the see-saw to be in balance:
[tex]\[ \text{Moment of Ram} = \text{Moment of Shyam} \][/tex]
[tex]\[ WR \times dR = WS \times dS \][/tex]
Where [tex]\( dS \)[/tex] is the distance Shyam should sit from the fulcrum.
[tex]\[ 400 \times 2 = 500 \times dS \][/tex]
Now solve for [tex]\( dS \)[/tex]:
[tex]\[ dS = \frac{400 \times 2}{500} \][/tex]
[tex]\[ dS = \frac{800}{500} \][/tex]
[tex]\[ dS = 1.6 \text{ meters} \][/tex]
Answer: Shyam should sit 1.6 meters from the fulcrum to balance Ram.
### Problem (c):
The mechanical advantage of a simple machine is 3 and its velocity ratio is 4. Find its efficiency.
Solution:
1. Mechanical Advantage (MA) is 3.
2. Velocity Ratio (VR) is 4.
The efficiency of a simple machine is given by:
[tex]\[ \text{Efficiency} (\%) = \left( \frac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} \right) \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = \left( \frac{3}{4} \right) \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = 0.75 \times 100 \][/tex]
[tex]\[ \text{Efficiency} (\%) = 75 \][/tex]
Answer: The efficiency of the machine is 75%.
