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Forces on an Object

  • #1

Homework Statement


https://s.yimg.com/hd/answers/i/b34ac93e70c94cf6a0bb58ff37d2a580_A.jpeg?a=answers&mr=0&x=1422893817&s=2f584c4d25bb63ccec270dfa3fcaa37e [Broken]

2. Question

How would one go about solving the forces acting at the supports in each of these structures?
The supports are the triangles. Anyone can give simple explanation on how to do these?
 
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Answers and Replies

  • #2
BvU
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Hello John, and welcome to PF :)

Aiming your phone at your lap and submitting it is not enough in the PF environment. You are supposed to fill in the template, possibly with a figure as a supportive illustration, and that means making an effort at solving the thing:

1. Homework Statement
2. Homework Equations
3. The Attempt at a Solution

And one problem at the time. Who knows, you understand what has to be done after a few nudges here and there, and then you can practice on the other seven exercises !

In the guidelines you can read why all these (mandatory) rules. It helps us help you at an appropriate level, too. In short: what do you know and what have you tried so far ?
 
  • #3
Hello John, and welcome to PF :)

In the guidelines you can read why all these (mandatory) rules. It helps us help you at an appropriate level, too. In short: what do you know and what have you tried so far ?
That the horizontal forces sum up to 0 if no forces are applied horizontally vice versa for vertical forces. That is all I know I am stuck from there on out. I have attempted doing the first part as follows:
The support on the left of part 1 is 2m away from the 10kN force therefore its value would be the vertical force applied divided by the distance away from the forces main point of contact. In other words: 10kN divided by 2m which is a 5kN force acting on the left support but I have been told this is not how you calculate it and I have no recommended text or notes to base this off. Also working in SI units is a challenge for me but its manageable. Thanks.
Could you perhaps guide me in layman's words on how one would approach calculating forces applied on the loads [triangles] ?
 
  • #4
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That the horizontal forces sum up to 0 if no forces are applied horizontally vice versa for vertical forces. That is all I know I am stuck from there on out. I have attempted doing the first part as follows:
The support on the left of part 1 is 2m away from the 10kN force therefore its value would be the vertical force applied divided by the distance away from the forces main point of contact. In other words: 10kN divided by 2m which is a 5kN force acting on the left support but I have been told this is not how you calculate it and I have no recommended text or notes to base this off. Also working in SI units is a challenge for me but its manageable. Thanks.
Could you perhaps guide me in layman's words on how one would approach calculating forces applied on the loads [triangles] ?
Have you learned that, for a rigid body to be in equilibrium, you must have a balance of forces and moments? Have you learned about moments yet?

Chet
 
  • #5
Have you learned that, for a rigid body to be in equilibrium, you must have a balance of forces and moments? Have you learned about moments yet?

Chet
Is that information necessary in calculating the forces applied on the supports? Is the questions I posted complex? I do not know how to do it as my attempts to achieving a right answer have failed so far but it seems like there should be and is a very simple explanation on how to go about the calculations. Would posting a picture of my attempt deem me worthy of a simple explanation to my problem? :)
 
  • #6
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Would posting a picture of my attempt deem me worthy of a simple explanation to my problem?
Yes :)
posting the attempt itself is even better: when you have to type it out your thinking has to be a little different, that might help.
 
  • #7
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Is that information necessary in calculating the forces applied on the supports?
Yes. So, again, have you learned how to do a balance of moments yet?

Chet
 
  • #8
BvU
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Could you perhaps guide me in layman's words on how one would approach calculating forces applied on the loads [triangles] ?
You can guess a little: in picture 1, forget about the forces and place yourself on the board. If you stand right on top of the left support, you expect that one to carry your weight. Idem right. And in the middle, your weight will be evenly distributed over the two.
Combine that with the knowledge that the sum of the forces the two supports provide is your weight.
Make a graph of force on right support as a function of x (distance to left support) : you have three points, what would be the relationship ?

And remember, Force divided by length doesn't yield a force ! Not in SI, but also not in any other system.
 
  • #9
Yes. So, again, have you learned how to do a balance of moments yet?

Chet
My tutor is incomprehensible so I suppose the answer would be a no yet I am expected to do this.
 
  • #10
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My tutor is incomprehensible so I suppose the answer would be a no yet I am expected to do this.
Do you have a text book? If so, what textbook are you using? (Your tutor is not going to be there to hold your hand when you get into the real world).

Chet
 
  • #11
Do you have a text book? If so, what textbook are you using? (Your tutor is not going to be there to hold your hand when you get into the real world).

Chet
Like I said, no textbook given and why state the obvious? I am in University to learn. The real world comes later, first comes acquiring and learning the skills to live in it.
 
  • #12
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Like I said, no textbook given and why state the obvious? I am in University to learn. The real world comes later, first comes acquiring and learning the skills to live in it.
Are you saying that you are taking a University course in Physics, and there is no textbook? Does your University have a library? What University is this?

Chet
 
  • #13
BvU
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He let's stick to the problem at hand ! To enroll in university, you have to have to fulfil some prerequisites, like college, primary school, or as an absolute minimum: kindergarten. So we're OK !

John: think see-saw in playground. Me at 100 kg, 2 m away from the support. Where does my kid brother (50 kg) have to sit on the other side to get equilibrium ?
 
  • #14
Are you saying that you are taking a University course in Physics, and there is no textbook? Does your University have a library? What University is this?

Chet
I am taking a course where the tutor does not give notes instead we must listen and write our own. No recommended texts were suggested by him. If you're not willing to help I'm sure I can find out off a class mate or something. I came here for the convenience and based on a friend's recommendation of the forum but I'm starting to think it is more inconvenient than helpful. I've asked this question well over an hour ago and instead of answers I've been receiving questions. I told what I've attempted, what my available resources are (i.e. not many) and have said would uploading a screenshot of my current workings be enough to qualify for help off somebody who knows how to do it.
I'll say it straight. I do not need somebody to do my work for me as I would learn nothing in the progress, however an explanation with examples would be extremely appreciated.
 
  • #15
He let's stick to the problem at hand ! To enroll in university, you have to have to fulfil some prerequisites, like college, primary school, or as an absolute minimum: kindergarten. So we're OK !

John: think see-saw in playground. Me at 100 kg, 2 m away from the support. Where does my kid brother (50 kg) have to sit on the other side to get equilibrium ?
This is my first year doing Physics, My Engineering course only required Mathematics to be accepted. As far as Physics is concerned it was recommended but not required.
Also to answer your question seeing how you are double his weight he would need to sit half the distance you are sitting from the point.
 
  • #16
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Fair enough. No more see-saw examples.
(Your post just came in). So you can accept that, to have equilibrium, F times distance on the left has to be equal to F times distance on the right, which is what I was aiming for.

F times distance is the torque. Just like Newton Σ F = 0 for equilibrium wrt translation, there is Σ torques = 0 to have equilibrium wrt angular motion. That's your second equation for figure 1.

Have to run now (I have a life too...)
 
  • #17
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I am taking a course where the tutor does not give notes instead we must listen and write our own. No recommended texts were suggested by him. If you're not willing to help I'm sure I can find out off a class mate or something. I came here for the convenience and based on a friend's recommendation of the forum but I'm starting to think it is more inconvenient than helpful. I've asked this question well over an hour ago and instead of answers I've been receiving questions. I told what I've attempted, what my available resources are (i.e. not many) and have said would uploading a screenshot of my current workings be enough to qualify for help off somebody who knows how to do it.
I'll say it straight. I do not need somebody to do my work for me as I would learn nothing in the progress, however an explanation with examples would be extremely appreciated.
As a moderator at Physics Forums, I'll say it straight also. We at Physics Forums are not here to teach you all the fundamentals of Physics; that should be covered in your course. We are only here to help you over the rough spots, and to help you to help yourself solve homework problems. You need to study the basics on your own. In posts #5 and #7, I mentioned moments and balance of moments. BvU also alluded to this. The way I will get started helping you is to suggest that you Google the subject Balance of Moments, and read the material in the links that are provided. If you have any questions about the material in these links, both BvU and I will be pleased to answer them. Once you gain some understanding of the material, we will also help you apply what you have learned to address the kinds of problems in post #1. Fair enough?

Chet
 
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  • #18
BvU
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Also to answer your question seeing how you are double his weight he would need to sit half the distance you are sitting from the point.
Oh boy, I misread this one and don't want to leave this uncorrected -- there may be others who read all this later on.

I am double his weight so he would need to sit twice the distance in order to have more leverage.

Try it out balancing a ruler with two erasers on one end and one on the other: where do you have to support it ?

That's also why burglars use the short end of a crowbar to pry between the door and the frame, and manhandle the long end...
 
  • #19
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How is John getting along ? 8 exercises to complete and in number 5 there isn't even a force in attendance !
 

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