A uniform steel bar of mass 5 kg and the length 1.5 m is supported at its two ends. A mass of 2 kg is suspended from apt 0.3m from the left end of the bar. Determine the reactions at each end of the steel bar.

To solve the problem of determining the reactions at each end of the steel bar, we can apply the principles of static equilibrium. In this scenario, the bar is supported at both ends, and we need to consider the forces acting on it, including the weight of the bar itself and the additional weight suspended from it. Let's break this down step by step.

Understanding the Forces Involved

First, we need to identify all the forces acting on the bar:

• The weight of the steel bar, which acts at its center of gravity.
• The weight of the suspended mass, which acts at a specific point along the bar.
• The reactions at the supports (let's call them R1 at the left end and R2 at the right end).

Calculating the Weights

The weight of the steel bar can be calculated using the formula:

Weight = mass × gravity

Given that the mass of the steel bar is 5 kg, and using the acceleration due to gravity as approximately 9.81 m/s²:

Weight of the bar = 5 kg × 9.81 m/s² = 49.05 N

Next, for the suspended mass of 2 kg:

Weight of the suspended mass = 2 kg × 9.81 m/s² = 19.62 N

Setting Up the Equilibrium Equations

For the bar to be in static equilibrium, the sum of the vertical forces must equal zero, and the sum of the moments about any point must also equal zero. We can set up our equations accordingly.

Sum of Vertical Forces

The equation for vertical forces can be expressed as:

R1 + R2 - Weight of the bar - Weight of the suspended mass = 0

Substituting the known values:

R1 + R2 - 49.05 N - 19.62 N = 0

Thus, we have:

R1 + R2 = 68.67 N

Sum of Moments

Next, we can take moments about one of the supports to find the reactions. Let's take moments about the left end (where R1 is located). The moment due to a force is calculated as the force multiplied by the distance from the pivot point.

The distance from the left end to the center of the bar (where its weight acts) is:

0.75 m (half of 1.5 m)

The distance from the left end to the suspended mass is:

0.3 m

The moment equation about the left end is:

R2 × 1.5 m - Weight of the bar × 0.75 m - Weight of the suspended mass × 0.3 m = 0

Substituting the weights:

R2 × 1.5 m - 49.05 N × 0.75 m - 19.62 N × 0.3 m = 0

Calculating the moments:

R2 × 1.5 m - 36.7875 N·m - 5.886 N·m = 0

Thus:

R2 × 1.5 m = 42.6735 N·m

Solving for R2:

R2 = 42.6735 N·m / 1.5 m = 28.449 N

Finding R1

Now that we have R2, we can substitute it back into the vertical forces equation to find R1:

R1 + 28.449 N = 68.67 N

Therefore:

R1 = 68.67 N - 28.449 N = 40.221 N

Final Results

In summary, the reactions at each end of the steel bar are:

• R1 (left end) = 40.221 N
• R2 (right end) = 28.449 N

This analysis shows how to apply the principles of static equilibrium to solve for unknown forces acting on a system. If you have any further questions or need clarification on any part of the process, feel free to ask!