A system of pulley has a velocity ratio of 6 and mechanical advantage of 5 the system is used to raise a load of 3000N through 10m calculate the workdone on load
Work done on the load = force applied x distance moved in the direction of force
In this case, the force applied is the effort force, which is the force required to lift the load, and the distance moved is the height through which the load is lifted.
The mechanical advantage of the pulley system is given by:
MA = Load / Effort
Since the MA is 5, the effort force required to lift the load is:
Effort = Load / MA
Effort = 3000N / 5
Effort = 600N
The velocity ratio of the pulley system is given by:
Velocity ratio = distance moved by effort / distance moved by load
Since the velocity ratio is 6, the distance moved by the effort force is:
Distance moved by effort = 6 x Distance moved by load
Distance moved by load = 10m (given)
Distance moved by effort = 6 x 10m
Distance moved by effort = 60m
Therefore, the work done on the load is:
Work done on load = Effort x Distance moved by load
Work done on load = 600N x 10m
Work done on load = 6000 Joules
Therefore, the work done on the load is 6000 Joules.
To calculate the work done on the load, we need to multiply the force applied to the system by the distance the load is raised.
The mechanical advantage (MA) is defined as the ratio of the output force (load) to the input force (applied force). In this case, the mechanical advantage is given as 5.
MA = Load / Applied force
Since the mechanical advantage is given and the load is known (3000N), we can calculate the applied force:
5 = 3000N / Applied force
Applied force = 3000N / 5
Applied force = 600N
Now we can calculate the work done on the load:
Work = Force × Distance
Work = 3000N × 10m
Work = 30000 N·m (or J - joules)