a non uniform bar which is 3m long weighs 60n. if the center of gravity is 1.1m from the larger end. and if the bar is supported at the midpoint, what load must be suspended at the smaller end to maintain equilibrium?
To find the load that must be suspended at the smaller end to maintain equilibrium, we need to consider the moments acting on the non-uniform bar.
The moment of a force is equal to the force multiplied by the perpendicular distance from the point of rotation (or support) to the line of action of the force. In this case, the point of rotation is at the midpoint of the bar.
Let's break down the problem into smaller steps:
Step 1: Calculate the weight of the bar
Given that the bar weighs 60 N, this weight acts at the center of gravity of the bar. Since the center of gravity is stated to be 1.1 m from the larger end, we can calculate the moment due to the weight of the bar. The moment (M_weight) is given by the weight (60 N) multiplied by the distance from the center of gravity to the midpoint (1.1 m). So, M_weight = 60 N * 1.1 m.
Step 2: Calculate the force required to maintain equilibrium
At equilibrium, the sum of the clockwise moments about the midpoint support and the anticlockwise moments about the midpoint support should be equal.
Let the load to be suspended at the smaller end be F_N (in newtons). The moment (M_support) due to the load F_N at the smaller end is given by: M_support = F_N * 1.5 m (since the distance from the smaller end to the midpoint is half the length of the bar).
Step 3: Equate the moments
Since the bar is in equilibrium, the clockwise and anticlockwise moments will cancel each other out. Thus, the moment due to the weight of the bar should be equal to the moment due to the load:
M_weight = M_support
Therefore, we have: 60 N * 1.1 m = F_N * 1.5 m
Simplifying:
66 N*m = F_N * 1.5 m
Step 4: Solve for F_N
Divide both sides of the equation by 1.5 m:
F_N = 66 N * 1.5 m / 1.5 m
F_N = 66 N
So, to maintain equilibrium, a load of 66 N must be suspended at the smaller end of the bar.