A uniform rod of weight 10N is balance at a point 75cm from the end B the pivot is removed to point 30cm from A. What force must be applied at A to balance the rod horizontally?
The fulcrum F is moved so that AF:FB = 1:4
So, the mass of AB is 1/4 the mass of FB
Now, since we have AF=30 and FB=120,
So, 8*30 = 2*120
To solve this problem, we can use the principle of moments, which states that in order for a body to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
In this case, the rod is balanced horizontally, meaning that the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let's start by calculating the clockwise moments. The weight of the rod, which acts at its center of gravity, creates a clockwise moment about point B. The weight is given as 10N and the distance from the center of gravity to point B is 75cm. So, the clockwise moment is given by:
Clockwise moment = weight * distance = 10N * 75cm
Now, let's calculate the anticlockwise moments. The force applied at point A creates an anticlockwise moment, as it is trying to rotate the rod counterclockwise. The distance from point A to the center of gravity is 30cm. Let's assume the force applied at point A is F.
Anticlockwise moment = force * distance = F * 30cm
Since the rod is balanced horizontally, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. Therefore:
Clockwise moment = Anticlockwise moment
10N * 75cm = F * 30cm
To find the force (F) required to balance the rod horizontally, we can rearrange the equation and solve for F:
F = (10N * 75cm) / 30cm
F = 25N
Therefore, to balance the rod horizontally, a force of 25N must be applied at point A.