To find the minimum effort required to raise the load using a block tackle, we need to consider the velocity ratio and the efficiency of the system.
The velocity ratio (VR) is defined as the ratio of the distance the effort moves to the distance the load moves. In this case, the velocity ratio is given as 4.
The efficiency of the system (η) is given as 80%, which can be written as 0.8.
We can use the formula for efficiency to calculate the actual effort required:
Efficiency (η) = (Output work / Input work) * 100%
Given that efficiency (η) = 0.8, we can rewrite the formula as:
0.8 = (Output work / Input work)
Since efficiency is defined as the ratio of output work to input work, we can rewrite the equation as:
Output work = 0.8 * Input work
The work done by the effort (Input work) is given by:
Input work = Effort * Distance moved by the effort
The work done against gravity by the load (Output work) is given by:
Output work = Load * Distance moved by the load
We know that the distance moved by the load is equal to the distance moved by the effort divided by the velocity ratio:
Distance moved by the load = Distance moved by the effort / Velocity ratio
Now, let's substitute the values given in the problem:
Load = 50 kg Velocity ratio (VR) = 4
Distance moved by the load = Distance moved by the effort / VR
The distance moved by the effort is the same as the distance moved by the load, so we can write:
Distance moved by the effort = Distance moved by the load = x (let's assume)
Now, substituting these values into the equations, we get:
Output work = Load * (Distance moved by the effort / VR) Input work = Effort * Distance moved by the effort
According to the efficiency formula:
Output work = 0.8 * Input work
Substituting the expressions for output work and input work, we have:
Load * (Distance moved by the effort / VR) = 0.8 * (Effort * Distance moved by the effort)
Simplifying the equation:
Load / VR = 0.8 * Effort
Effort = (Load / VR) / 0.8
Now, let's substitute the values:
Load = 50 kg VR = 4 Effort = (50 / 4) / 0.8 = 15.625 kg
Therefore, the minimum effort required to raise the load is approximately 15.625 kg.