A pulley system reduces the effort of lifting a big bucket of water from a shallow well. The 100-newton bucket is lifted to a height of 5 meters with an unknown input force. If 50 meters of rope is pulled to lift the bucket the desired height and the percent efficiency of the pulley system is 78%, what must the input force be?
To find the input force required to lift the bucket using the pulley system, we can use the formula for mechanical efficiency:
Efficiency = (Output force / Input force) * 100
Given that the efficiency of the pulley system is 78%, we can substitute this value into the formula:
78 = (100 / Input force) * 100
To isolate the input force, we can rearrange the formula:
Input force = (100 / Efficiency) * 100
Now, we can substitute the given efficiency value and solve for the input force:
Input force = (100 / 78) * 100
Input force = 128.21 Newtons (rounded to two decimal places)
Therefore, the input force required to lift the bucket using the pulley system is approximately 128.21 Newtons.
To solve this problem, we need to use the concept of work and efficiency.
Work is defined as the force applied over a certain distance. In this case, the work done to lift the bucket can be calculated using the formula:
Work = Force * Distance
In the given problem, the force required to lift the bucket is unknown, let's call it "F". The distance over which the force is applied is 50 meters (the length of the rope that is pulled). The work done to lift the bucket is the product of the force and the distance:
Work = F * 50
Next, we can calculate the work done against gravity to lift the bucket to a height of 5 meters. The work done against gravity is given by the formula:
Work against gravity = Force * Height
In this case, the force is 100 Newtons (the weight of the bucket) and the height is 5 meters. The work done against gravity is:
Work against gravity = 100 * 5 = 500
Now, we can calculate the efficiency of the pulley system. Efficiency is defined as the ratio of the useful work output to the work input, expressed as a percentage:
Efficiency = (Useful work output / Work input) * 100
In this case, the useful work output is the work done against gravity (500) and the work input is the work done to lift the bucket (F * 50). The efficiency is given as 78% (0.78).
0.78 = (500 / (F * 50)) * 100
To solve for F, we can rearrange the equation:
F * 50 = 500 / 0.78
F = (500 / 0.78) / 50
Using a calculator, we can determine the value of F:
F ≈ 6.4102
Therefore, the input force required to lift the bucket is approximately 6.4102 Newtons.