A crowbar ,1.5m long is used to lift a 100kg rock out of a hole 25cm deep.If the crowbar pivots 50cm from one end ,what is the force that must be used to just lift the rock ?
To find the force required to lift the rock using the crowbar, we can use the principle of moments. The principle of moments states that the sum of the moments acting on an object in equilibrium is zero.
Let's break down the problem step by step:
1. Determine the weight of the rock:
The weight of the rock is given as 100 kg. We can calculate the force exerted by gravity using the formula: Weight = mass × acceleration due to gravity. Taking the acceleration due to gravity as approximately 9.8 m/s^2, the weight of the rock is 100 kg × 9.8 m/s^2 = 980 Newtons (N).
2. Determine the distance from the pivot to the weight:
The distance from the pivot to the weight is 25 cm, which we need to convert to meters. 25 cm = 0.25 m.
3. Determine the distance from the pivot to the point where the force is applied:
The distance from the pivot to the point where the force is applied is 50 cm, which is also 0.5 m.
Now, let's apply the principle of moments:
The clockwise moment = Force × Distance from pivot to force
The anti-clockwise moment = Weight × Distance from pivot to weight
In equilibrium, the sum of the clockwise moments and the anti-clockwise moments is zero.
So, Force × 0.5 m = 980 N × 0.25 m
Now we can solve for Force:
Force = (980 N × 0.25 m) / 0.5 m
Force = 490 Newtons (N)
Therefore, the force required to just lift the rock using the crowbar is 490 Newtons.