what is the minimum effort required to raise a load of 5kg in an ideal pulley system taking gravity to be 10m/s/s
In ideal pulley, Velocity Ratio = 1
Effort Required = Weight x Velociy Ratio = mass x acc. due to gravity x 1 = 5 x 10 x 1 = 50 Newton.
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To determine the minimum effort required to raise a load of 5kg in an ideal pulley system, we need to understand the concept of mechanical advantage.
In an ideal pulley system, the mechanical advantage is equal to the number of supporting ropes or segments of rope that support the load. In this case, if we assume a simple system with a single movable pulley, the mechanical advantage is 2. This means that for every unit of effort applied, the load is raised by 2 units.
Using this information, we can calculate the minimum effort required to lift the load.
First, let's determine the weight of the load. The weight of an object is given by the formula:
Weight = mass x acceleration due to gravity
In this case, the mass is 5kg and the acceleration due to gravity is 10m/s^2. So, the weight of the load is:
Weight = 5kg x 10m/s^2 = 50 N (Newton)
To lift this load, we need a force equal to the weight of the load. However, since the mechanical advantage is 2, the effort required will be half of the weight of the load.
Effort = Weight / Mechanical Advantage
= 50N / 2
= 25N
Therefore, the minimum effort required to raise the load of 5kg in an ideal pulley system is 25 Newtons.