Moment Question / inclined triangle

In summary: Then I resolved it to 2cos(30). In summary, you found the weight of the 12kg mass, found the moment due to the weight of the bonnet, and found the tension due to the moment.
  • #1
Sabeshan Ratneswaran
18
1

Homework Statement


In this I tried to resolve the components.

So first thing, I converted the 12kg into Newtons so it would be 117.72 Newtons. Then found the perpendicular distance which is to g: cos(30)x1 then multiply the answer by 117.2N to give the weight down as 102N. As the moment principle state that clockwise have to be equal to anticlockwise the vertical of the Q have to 102N then i found the tension by 102/cos(30) which will give the final answer as 118N. But this answer is not one of the choices so it is wrong. Can someone explain to me where I have gone wrong.

Thank You.
 

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  • #2
Sabeshan Ratneswaran said:
o give the weight down as 102N.
No, the weight down is mg. In what direction is there a force mg cos(30)?
To balance the moments, pick an axis. Which axis did you use?
 
  • #3
Sabeshan Ratneswaran said:
So first thing, I converted the 12kg into Newtons so it would be 117.72 Newtons.
You found the weight of the 12kg mass. Good.

Sabeshan Ratneswaran said:
Then found the perpendicular distance which is to g: cos(30)x1 then multiply the answer by 117.2N to give the weight down as 102N.
You found the anticlockwise torque (moment) due to the weight of that bonnet. Good. (Not the "weight" down, the moment.)

Sabeshan Ratneswaran said:
As the moment principle state that clockwise have to be equal to anticlockwise
Good.

Sabeshan Ratneswaran said:
then i found the tension by 102/cos(30)
Not sure what you did here. What you need to do is find an expression for the clockwise moment due to the force F. (What's the definition of moment?)
 
  • #4
Doc Al said:
You found the weight of the 12kg mass. Good.You found the anticlockwise torque (moment) due to the weight of that bonnet. Good. (Not the "weight" down, the moment.)Good.Not sure what you did here. What you need to do is find an expression for the clockwise moment due to the force F. (What's the definition of moment?)
Hi there, This is where I am confused. Moment about a pivot is force multiplied by the perpendicular distance between the the pivot and the line of action
 
  • #5
We know the vertical component as 102 N so we can resolve to find the tension. This is where I got the expression as 102/cos(30)
 
  • #6
Sabeshan Ratneswaran said:
We know the vertical component as 102 N so we can resolve to find the tension. This is where I got the expression as 102/cos(30)
No, the vertical force is mg, 117.72N.
You can either take the component of that which is perpendicular to the car bonnet (that will be 102N) and multiply by the distance from the pivot to that force; or take the whole of mg and multiply by the perpendicular (i.e. horizontal) distance from the pivot to the line of action of mg.
You will get the same answer either way. Choose one.
 
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  • #7
haruspex said:
No, the vertical force is mg, 117.72N.
You can either take the component of that which is perpendicular to the car bonnet (that will be 102N) and multiply by the distance from the pivot to that force; or take the whole of mg and multiply by the perpendicular (i.e. horizontal) distance from the pivot to the line of action of mg.
You will get the same answer either way. Choose one.

Quite confused:

is this what you mean 12x9.81xcos(30)x1=102N
Then divide it by cos(30)x2 to give 58.86.

but we are trying to find out the force of F which holds the bonnet equilibrium. By the way the answer is not 59N.
 
  • #8
Sabeshan Ratneswaran said:
Then divide it by cos(30)x2 to give 58.86
Explain why you divide by that.
 
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  • #9
haruspex said:
Explain why you divide by that.
The principle of moment is the sum of clockwise moment is equal to sum of anticlockwise moment in an equilibrium position. the expression will be FxD=FxD
So basically we are dealing with perpendicular distance so the expression with be [Cos(30)x1] x [9.81x12]=[Cos(30)x2] x F
So making the F the subject will give us 58.86N.
 
  • #10
haruspex said:
Explain why you divide by that.

I am only dealing with resolved distance components
 
  • #11
Sabeshan Ratneswaran said:
The principle of moment is the sum of clockwise moment is equal to sum of anticlockwise moment in an equilibrium position. the expression will be FxD=FxD
So basically we are dealing with perpendicular distance so the expression with be [Cos(30)x1] x [9.81x12]=[Cos(30)x2] x F
So making the F the subject will give us 58.86N.
Yes, I understand that, but how do you come to 2 cos(30)? What is the perpendicular distance from the line of action of F to the pivot point?
 
  • #12
haruspex said:
Yes, I understand that, but how do you come to 2 cos(30)? What is the perpendicular distance from the line of action of F to the pivot point?
It is not from P to F, the distant from P to Q is 2 so resolved it do 2cos(30)..

then I was going to find the value of F by doing 58.86/ cos(30). But i know I have done something wrong.
 
  • #13
Sabeshan Ratneswaran said:
It is not from P to F, the distant from P to Q is 2 so resolved it do 2cos(30)..
I'll be more specific. Why the cos(30)?
 
  • #14
Sabeshan Ratneswaran said:

Homework Statement


In this I tried to resolve the components.

So first thing, I converted the 12kg into Newtons so it would be 117.72 Newtons. Then found the perpendicular distance which is to g: cos(30)x1 then multiply the answer by 117.2N to give the weight down as 102N. As the moment principle state that clockwise have to be equal to anticlockwise the vertical of the Q have to 102N then i found the tension by 102/cos(30) which will give the final answer as 118N. But this answer is not one of the choices so it is wrong.Can someone explain to me where I have gone wrong.

Thank You.
haruspex said:
I'll be more specific. Why the cos(30)?
 

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  • #15
Realize that the applied force F is already perpendicular to the distance PQ, so there is no need to take any components.
 
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  • #16
Doc Al said:
Realize that the applied force F is already perpendicular to the distance PQ, so there is no need to take any components.
Make sense mentor
 
  • #17
Sabeshan Ratneswaran said:
Make sense mentor
Look at the diagram. At what angle does F act? What is the perpendicular distance from the pivot point to the line of action of F?
 
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  • #18
F1 * d1 = F2 * d2 ?

bonnet forces.jpeg
 
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  • #19
haruspex said:
Look at the diagram. At what angle does F act? What is the perpendicular distance from the pivot point to the line of action of F?
Thank you for the consistent help.
 

FAQ: Moment Question / inclined triangle

1. What is a moment question?

A moment question is a problem that involves finding the force needed to balance or move an object that is resting on an inclined triangle. It requires knowledge of basic physics concepts such as forces, moments, and angles.

2. How do I solve a moment question involving an inclined triangle?

To solve a moment question involving an inclined triangle, you will need to use the principles of statics, which state that an object will remain in equilibrium if the sum of all forces and moments acting on it is equal to zero. You will also need to use trigonometry to calculate the angles and forces involved.

3. What information do I need to solve a moment question involving an inclined triangle?

You will need to know the weight of the object resting on the triangle, the angle of the incline, and the location of the force or forces acting on the object. You may also need to know the coefficients of friction and the weight distribution of the object.

4. Can you give an example of a moment question involving an inclined triangle?

Sure, an example of a moment question involving an inclined triangle would be: "A 10 kg box is resting on an inclined triangle with an angle of 30 degrees. The weight of the box is evenly distributed. What is the minimum force needed to push the box up the incline with a constant velocity?"

5. What are some real-life applications of moment questions involving inclined triangles?

Moment questions involving inclined triangles are commonly used in engineering and construction to calculate the stability and strength of structures such as bridges, ramps, and roofs. They are also used in physics experiments to study the effects of forces and friction on objects.

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