Equivalent Resultant Force for Distributed Load: Is My Method Correct?

In summary, the problem involves finding the equivalent resultant force for a distributed load by dividing it into four parts. However, there are discrepancies in the given information, such as a 5 kN concentrated load at point D and a 5 kN-m torque at point C, which may affect the accuracy of the calculations.
  • #1
goldfish9776
310
1

Homework Statement


i am asked to find the equivalent resultant force for this question . for the distributed load , i divided into 4 parts , is my method correct?

Homework Equations

The Attempt at a Solution


total force about y = 6sin30 + 3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5)

total force about x = 6cos 30 [/B]
 

Attachments

  • DSC_0365.JPG
    DSC_0365.JPG
    27.2 KB · Views: 444
Physics news on Phys.org
  • #2
goldfish9776 said:
3. The Attempt at a Solution
total force about y = 6sin30 + 3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5)

total force about x = 6cos 30
What do you mean by "about x" etc.? You seem to be confusing forces and moments. The 6 sin 30 is a force in the y direction, but the 3(3) looks more like a moment, though I can't match it to anything in the diagram (there is no 3kN force anywhere), and the 5(4) might be the moment of the 5kN force at D about E.
Please identify what each term represents.
 
  • #3
haruspex said:
What do you mean by "about x" etc.? You seem to be confusing forces and moments. The 6 sin 30 is a force in the y direction, but the 3(3) looks more like a moment, though I can't match it to anything in the diagram (there is no 3kN force anywhere), and the 5(4) might be the moment of the 5kN force at D about E.
Please identify what each term represents.
3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5) , each term here represent the different region for the load distribution diagram , i already stated it in the diagram
sorry , i mean forec along x and y axis
 
  • #4
goldfish9776 said:
3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5) , each term here represent the different region for the load distribution diagram , i already stated it in the diagram
sorry , i mean forec along x and y axis
You did indeed, but I still cannot see how you are getting these terms.
First, is this equation supposed to be for forces in the x direction, forces in the y direction, or moments about some point? If moments about a point, what point?
Secondly, how are you getting 3(3)? I cannot relate that to anything in the diagram. There is no force of 3kN. Perhaps this stands for the total force from the 3kN/m over the 3m from C to D, but that doesn't make sense because the force density increases from 3kN/m at C to 33/7kN/m at D.
Thirdly, where does the (5-3) come from? Are the 5 and the 3 distances or forces?
The 5(4) looks like the moment of the 5kN force at D about E. Please confirm.
I can't make sense of the 7-5 term either.
 
  • #5
goldfish9776 said:

Homework Statement


i am asked to find the equivalent resultant force for this question . for the distributed load , i divided into 4 parts , is my method correct?

Homework Equations

The Attempt at a Solution


total force about y = 6sin30 + 3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5)

total force about x = 6cos 30 [/B]
You really should learn the formula for calculating the area of a trapezoid:

images?q=tbn:ANd9GcRx59eXAyPKKQtPsgXNpdkr19ScTCdCGfX46gqjHw5kz9gwhxXZMA.jpg

It saves quite a bit of time. If b1 or b2 is zero, this formula reduces to that of finding the area of a triangle, A = (1/2)hb.

You should be aware that the load shown at point D is a 5 kN concentrated load. It is a separate load from the distributed load.

The value of the distributed load at point D is not 5 kN/m. The distributed loading changes by 4 kN/m over a distance of 7 m, so the value of the distributed loading at point D is calculated by linear interpolation. This step is unnecessary if you know how to calculate the area of a trapezoid, as shown above.

There is also a couple of magnitude 5 kN-m acting CCW shown at point C, which doesn't seem to appear in your calculations.

The problem statement as given is confusing, as haruspex noted. Please provide the exact wording for this problem from your text.
 
  • #6
SteamKing said:
You really should learn the formula for calculating the area of a trapezoid:

images?q=tbn:ANd9GcRx59eXAyPKKQtPsgXNpdkr19ScTCdCGfX46gqjHw5kz9gwhxXZMA.jpg

It saves quite a bit of time. If b1 or b2 is zero, this formula reduces to that of finding the area of a triangle, A = (1/2)hb.

You should be aware that the load shown at point D is a 5 kN concentrated load. It is a separate load from the distributed load.

The value of the distributed load at point D is not 5 kN/m. The distributed loading changes by 4 kN/m over a distance of 7 m, so the value of the distributed loading at point D is calculated by linear interpolation. This step is unnecessary if you know how to calculate the area of a trapezoid, as shown above.

There is also a couple of magnitude 5 kN-m acting CCW shown at point C, which doesn't seem to appear in your calculations.

The problem statement as given is confusing, as haruspex noted. Please provide the exact wording for this problem from your text.

so the total forecs along y should be = 3(7) + 5 + 0.5(4)(7 ) ? , since you aid that 5N is concentrated load...
 
  • #7
goldfish9776 said:
so the total forecs along y should be = 3(7) + 5 + 0.5(4)(7 ) ? , since you aid that 5N is concentrated load...
Yes, though I would have done it as 5+(3+4)(3+7)/2, the (3+7)/2 being the average spread load.
SteamKing mentioned the 5kNm torque, but since that is a pure torque it does not feature in the net force expressions.
 

Related to Equivalent Resultant Force for Distributed Load: Is My Method Correct?

1. What is the concept of equivalent resultant force?

The concept of equivalent resultant force is a principle in physics that states that multiple forces acting on an object can be replaced by a single force that has the same effect on the object's motion.

2. How is equivalent resultant force calculated?

Equivalent resultant force is calculated by adding together all of the individual forces acting on an object using vector addition. This results in a single force vector that represents the combined effect of the multiple forces.

3. What is the difference between equivalent resultant force and net force?

Equivalent resultant force and net force are often used interchangeably, but there is a subtle difference between the two. Equivalent resultant force refers to a single force that represents the combined effect of multiple forces, while net force specifically refers to the sum of all the forces acting on an object in a specific direction.

4. How does equivalent resultant force affect an object's motion?

Equivalent resultant force determines the direction and magnitude of an object's motion. If the equivalent resultant force is zero, the object will remain at rest or continue moving at a constant velocity. If the equivalent resultant force is non-zero, the object will accelerate in the direction of the force.

5. Can equivalent resultant force be negative?

Yes, equivalent resultant force can be negative. This occurs when the individual forces acting on an object are in opposite directions, resulting in a net force that is in the opposite direction. In this case, the negative equivalent resultant force will cause the object to accelerate in the opposite direction of the net force.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
314
  • Introductory Physics Homework Help
Replies
1
Views
873
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
Back
Top