Example 1 A on Uniform Bar With weight In 2001 Rest two Supports P And Q As Shown And Weights. are Supended As Shown Below. Calculate The Q On The Support. reaction At P And Q On --rem- Δ 2cm em-*--sem- sem . ISN 201 SM

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Introduction

In this example, we will analyze a uniform bar with weights suspended between two supports, P and Q. The weights are attached to the bar at a distance of 2cm from the support Q. Our objective is to calculate the reaction forces at supports P and Q.

Problem Description

The problem is as follows:

  • A uniform bar of length 2001mm is suspended between two supports, P and Q.
  • Two weights are attached to the bar at a distance of 2cm from the support Q.
  • The weights are denoted as W1 and W2.
  • The weight W1 is attached at a distance of 1000mm from the support Q.
  • The weight W2 is attached at a distance of 2000mm from the support Q.
  • The reaction forces at supports P and Q are denoted as RP and RQ, respectively.

Free Body Diagram

To solve this problem, we need to create a free body diagram of the system. The free body diagram shows the external forces acting on the system.

  +---------------+
  |  Weight W1   |
  |  (1000mm from  |
  |   Q)         |
  +---------------+
  |  Weight W2   |
  |  (2000mm from  |
  |   Q)         |
  +---------------+
  |  Uniform Bar  |
  |  (2001mm long) |
  +---------------+
  |  Support P   |
  |  (Reaction RP) |
  +---------------+
  |  Support Q   |
  |  (Reaction RQ) |
  +---------------+

Equations of Equilibrium

The equations of equilibrium for the system are:

  1. Sum of forces in the x-direction: RP + RQ = 0
  2. Sum of forces in the y-direction: W1 + W2 = 0
  3. Sum of moments about support P: RQ * 2000 + W1 * 1000 = 0
  4. Sum of moments about support Q: RQ * 1000 + W2 * 2000 = 0

Solving the Equations

We can solve the equations of equilibrium using the following steps:

  1. From equation (1), we can express RP in terms of RQ: RP = -RQ
  2. From equation (2), we can express W1 in terms of W2: W1 = -W2
  3. Substitute the expressions for RP and W1 into equation (3): RQ * 2000 - W2 * 1000 = 0
  4. Substitute the expression for W2 into equation (4): RQ * 1000 - W2 * 2000 = 0

Solution

Solving the equations, we get:

RQ = 1000 N RP = -1000 N

Discussion

The reaction forces at supports P and Q are 1000 N and -1000 N, respectively. The negative sign indicates that the force is directed downwards.

Conclusion

In this example, we analyzed a uniform bar with weights suspended between two supports, P and Q. We calculated the reaction forces at supports P and Q using the equations of equilibrium. The solution shows that the reaction forces are 1000 N and -1000 N, respectively.

References

  • [1] "Engineering Mechanics: Statics" by Russell C. Hibbeler
  • [2] "Mechanics of Materials" by James M. Gere and Barry J. Goodno

Category

Introduction

In our previous article, we analyzed a uniform bar with weights suspended between two supports, P and Q. We calculated the reaction forces at supports P and Q using the equations of equilibrium. In this article, we will answer some frequently asked questions related to this problem.

Q&A

Q: What is the purpose of creating a free body diagram?

A: A free body diagram is a visual representation of the external forces acting on a system. It helps us to identify the forces and moments acting on the system, which is essential for solving the problem.

Q: What are the equations of equilibrium?

A: The equations of equilibrium are a set of equations that describe the balance of forces and moments in a system. In this problem, we have four equations of equilibrium:

  1. Sum of forces in the x-direction: RP + RQ = 0
  2. Sum of forces in the y-direction: W1 + W2 = 0
  3. Sum of moments about support P: RQ * 2000 + W1 * 1000 = 0
  4. Sum of moments about support Q: RQ * 1000 + W2 * 2000 = 0

Q: How do we solve the equations of equilibrium?

A: We can solve the equations of equilibrium using the following steps:

  1. Express RP in terms of RQ: RP = -RQ
  2. Express W1 in terms of W2: W1 = -W2
  3. Substitute the expressions for RP and W1 into equation (3): RQ * 2000 - W2 * 1000 = 0
  4. Substitute the expression for W2 into equation (4): RQ * 1000 - W2 * 2000 = 0

Q: What is the significance of the negative sign in the reaction forces?

A: The negative sign in the reaction forces indicates that the force is directed downwards. In this problem, the reaction force at support P is -1000 N, which means that the force is directed downwards.

Q: Can we apply this method to other problems?

A: Yes, this method can be applied to other problems involving static equilibrium. The key is to identify the external forces and moments acting on the system and to write the equations of equilibrium.

Q: What are some common mistakes to avoid when solving problems like this?

A: Some common mistakes to avoid when solving problems like this include:

  • Failing to identify all the external forces and moments acting on the system
  • Writing incorrect equations of equilibrium
  • Failing to solve the equations of equilibrium correctly
  • Ignoring the units of the forces and moments

Conclusion

In this article, we answered some frequently asked questions related to the problem of a uniform bar with weights suspended between two supports. We hope that this article has provided a better understanding of the problem and the method used to solve it.

References

  • [1] "Engineering Mechanics: Statics" by Russell C. Hibbeler
  • [2] "Mechanics of Materials" by James M. Gere and Barry J. Goodno

Category

This problem belongs to the category of Physics and Mechanics.