an effort of 250N is applied to a pulley system consisting of two fixed and two movable pulleys to lift a load to a height of 20m. if the efficiency of the system is 60% and the work done in overcoming friction is 3000j, calculate
i) the mechanical advantage of the system
ii) the weight lifted
iii) the work done by the effort
Answer
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To solve this problem, we'll first calculate the values step by step.
i) The mechanical advantage of the system:
The mechanical advantage (MA) of a pulley system is calculated by dividing the load force by the effort force. Since the efficiency of the system is given, we can calculate the ideal (theoretical) load force by dividing the effort force by the efficiency. In this case, the effort force is 250N and the efficiency is 60%.
To calculate the ideal load force (F_load_ideal):
F_load_ideal = effort force / efficiency
F_load_ideal = 250N / 0.60
F_load_ideal = 416.67N
Since we have two movable pulleys in the system, the MA is equal to twice the number of movable pulleys, which is 4 in this case.
Therefore, the mechanical advantage (MA) of the system is:
MA = 4
ii) The weight lifted:
The weight lifted refers to the force required to lift the load. Therefore, the weight lifted is equal to the load force. Since the efficiency was accounted for when calculating the ideal load force, we can simply use the ideal load force as the weight lifted.
The weight lifted is:
Weight lifted = F_load_ideal
Weight lifted = 416.67N
iii) The work done by the effort:
The work done by the effort is calculated as the product of the effort force and the distance the effort force acts through. In this case, the distance is given as 20m.
The work done by the effort is:
Work done by effort = effort force * distance
Work done by effort = 250N * 20m
Work done by effort = 5000J
Therefore, the answers to the calculations are:
i) The mechanical advantage of the system is 4.
ii) The weight lifted is 416.67N.
iii) The work done by the effort is 5000J.