Solving for Vertical and Horizontal Forces in a Statics Problem

  • Thread starter peleus
  • Start date
  • Tags
    Statics
In summary, the conversation involves a person seeking help with a physics problem involving finding the vertical and horizontal components of a force using the sum of moments. They receive a hint to use the formula Torque = rFsinθ and eventually find the correct answer with the help of another individual confirming their working.
  • #1
peleus
18
0

Homework Statement



http://home.exetel.com.au/peleus/EIDEprob.jpg


2. The attempt at a solution

Basically did the standard sum of moments around point A to try and get the vertical component of point D.

-6.2*997 + 9.2*Dy - 12.4*1472

Dy = 2655.89

From there working out Dx

Dx = 2655.89 / tan 50

Dx = 2228.56

Then doing some Pythagoras finding CD to be 3467N.

Supplied answer is 2476N.


I think I'm going wrong for a few possible reasons,

1 - It's possible that the distances I'm using aren't perpendicular to the forces, I tried working out the perpendicular forces however and got a very similar answer.

2 - The weight of the bar and weight on the end act straight down, so I might need to somehow figure out a component of their force, but I have no idea how.

If anyone could show me the working for the correct answer for this, it would be much appreciated.

Cheers.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
peleus said:
Basically did the standard sum of moments around point A to try and get the vertical component of point D.

-6.2*997 + 9.2*Dy - 12.4*1472
You didn't take the angle of the beam into consideration when finding the torques due to the weights.

Also, if the tension in the cable is T, then the component perpendicular to the beam is Tsin50.

Hint: Use Torque = rFsinθ to find the magnitude of the torques.
 
  • #3
I haven't learned particular form yet.

Any chance you can show me how to do it so I can learn by your working?

I completely understand the idea behind hinting along so someone can work it out for themselves but I think it will just be easier if I can see someone elses working so I'll know what to do with all the future examples.

Cheers.
 
  • #4
peleus said:
I haven't learned particular form yet.
You can always just find the component of the weight perpendicular to the beam. Find the angle that the beam makes with the vertical. If that angle is θ, then the component of weight perpendicular to the beam is Wsinθ. (Thus the torque due to the weight = rWsinθ.)
 
  • #5
Edit - Wrong, redoing working.

Ok, if someone can confirm this for me I'll be forever in your debt.

Using the sine rule we find angle ACD to be 57 deg.

This leaves angle CAD to be 73 deg. This also means the beam is 17 deg above the horizon.

The result of this is that the weight's have 73 (72.9652 to be precise) degrees of angle between them and the weights.

--- How the textbook go their answer ---

Sum of moments around point A, counter-clockwise is positive.

9.2 * Dy - 1471.5 * 12.4 * sin72.9652
Dy = 1896.31
Dx = 1591.19

D resultant is 2575.45N (probably lost 0.05N in a rounding error somewhere to their 2576N answer).

--- The textbook didn't use the weight figure ---

Sum of moments around point A, counter-clockwise is positive.

-6.2 * 977.5 * sin72.9652 + 9.2 * Dy - 12.4 * 1471.5 * sin72.9652
Dy = 2526.16
Dx = 2119.70

D resultant is 3297.67N

If anyone could confirm my working is sound, you will avert me having a nervous breakdown.

I hate wrong answers.

Thank you.
 
Last edited:
  • #6
peleus said:
Using the sine rule we find angle ACD to be 57 deg.

This leaves angle CAD to be 73 deg. This also means the beam is 17 deg above the horizon.

The result of this is that the weight's have 73 (72.9652 to be precise) degrees of angle between them and the weights.
Looks good.

--- How the textbook go their answer ---

Sum of moments around point A, counter-clockwise is positive.

9.2 * Dy - 1471.5 * 12.4 * sin72.9652
Dy = 1896.31
Dx = 1591.19

D resultant is 2575.45N (probably lost 0.05N in a rounding error somewhere to their 2576N answer).

--- The textbook didn't use the weight figure ---
Looks like the book left out the weight of the beam. Oops!

Sum of moments around point A, counter-clockwise is positive.

-6.2 * 977.5 * sin72.9652 + 9.2 * Dy - 12.4 * 1471.5 * sin72.9652
Dy = 2526.16
Dx = 2119.70

D resultant is 3297.67N

If anyone could confirm my working is sound, you will avert me having a nervous breakdown.

I hate wrong answers.
Looks good to me. (Save a bit of effort by using Dy = D sin50.)
 
  • #7
Thank you very much for your help.
 

Related to Solving for Vertical and Horizontal Forces in a Statics Problem

1. What is statics?

Statics is a branch of mechanics that deals with the study of objects at rest or in a state of constant motion. It involves analyzing the forces acting on an object and their effects on its equilibrium.

2. What are the basic principles of statics?

The basic principles of statics are the law of inertia, the law of acceleration, and the law of action and reaction (also known as Newton's Laws of Motion). These principles help describe the behavior of objects at rest or in a state of constant motion.

3. How are forces represented in statics?

Forces in statics are typically represented by vectors, which have both magnitude and direction. They can be drawn as arrows, with the length of the arrow representing the magnitude of the force and the direction of the arrow representing the direction of the force.

4. What is the difference between a balanced and unbalanced force?

A balanced force refers to a system where the net force on an object is equal to zero, resulting in the object remaining at rest or moving at a constant velocity. An unbalanced force, on the other hand, results in a change in the object's motion, such as accelerating or decelerating.

5. How is statics applied in real-life situations?

Statics has many practical applications in everyday life, such as in the design and construction of buildings, bridges, and other structures. It is also used in the analysis of machines and vehicles, as well as in the study of fluid dynamics. Understanding statics can also help in predicting and preventing accidents or failures in various systems.

Similar threads

Engineering Solve X-Forces for Frame Statics Problem
    • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
Engineering Two approaches in statics not adding up
    • Engineering and Comp Sci Homework Help
Replies
5
Views
845
Statics Homework: Finding Tension and Force in a Walkway Supported by Rollers
    • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
Engineering Struggling with a Structural Analysis Problem?
    • Engineering and Comp Sci Homework Help
Replies
1
Views
880
Virtual Work and Static Equilibrium
    • Introductory Physics Homework Help
Replies
3
Views
876
I Statics: When the reactions depend on the displacements
    • Mechanics
Replies
5
Views
876
Body forces in static equilibrium (FEM issue)
    • Mechanical Engineering
Replies
2
Views
1K
Fluid Statics: Find vertical and horizontal forces in tank
    • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
Book plots position & force on the same graph, still gets right answer
    • Introductory Physics Homework Help
Replies
8
Views
226
Mastering 3D Statics: How to Solve Moments and Forces for Bearings A and B
    • Engineering and Comp Sci Homework Help
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top