What Are the Forces on a Supported Beam?

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The discussion centers on analyzing the forces acting on a supported beam with two students and a beam of known weight. Participants emphasize the importance of applying Newton's laws and the concept of torque to solve for the upward forces exerted by the supports. The challenge arises from the unknown weight of one student, which prevents a complete solution. Participants suggest using equilibrium conditions—sum of vertical forces and moments—to derive equations, but ultimately conclude that without the second student's weight, the problem cannot be solved. The conversation highlights the necessity of complete information for solving physics problems involving static equilibrium.
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Homework Statement



A 120 kg beam is 12.0 m long and has 2 supports, each of which are placed 2.0 m from each end (leaving 8.0 m of the beam between the 2 supports). A 50 kg student stands at the extreme left end of the beam, 2.0 m left of the left support. A second student stands 1.0 m from the right end which is 1.0 m to the right of the right support.

a. How much upward force does the right support exert on the beam?

b. How much upward force does the left support exert on the beam?


Homework Equations



F= ma
Torque?

The Attempt at a Solution



I found that each support holds 588 N due to the weighted rod but I am not sure how to account for the people and their distances from the supports.
 
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You need your definition of torque and you need to decide if the system is moving. If it is or it isn't, what information does this give you?

In other words what is the acceleration of the beam. You also need Newton's second law of rotational motion and you need to decide if the beam is rotating or not and what information that gives you.
 
T=dF, system is not moving, I am not sure what information this gives me, acceleration of the beam is 9.8 due to gravity?
 
Wikipedia has a definition of torque. Look it up. And yes it's a formula
The beam is resting on supports. Acceleration requires a change in velocity. If the beam is resting, is it moving? No. At rest means v = zero, change in v = zero since it stays at rest or motionless so it has no rotational motion either.

You need Newton's second law of rotational motion as well as linear motion which you have already given which is F = ma so you are on the right track.
You just need to draw your forces and distances and use the two formulas for Newton's second law - one with forces, the other with torques.
 
Newtons second law says that the force on the pivot is the distance from the pivot * F. Therefore does that mean on the left support it would be 2 * 50 * 9.8 = 980N on left support? but there is no weight given for the second student?
 
I think this is a question about a beam in equilibrium and the weight of the second student is not known.
If this is the case then there are 2 conditions that must be satisfied for equilibrium
1) There is no resultant vertical or horizontal force. This means that the sum of the forces (F1 and F2) at the supports must equal the total weight of the 2 students plus the beam itself.

2) There is no resultant turning effect/moment/torque (I believe that moment is the preferred term in equilibrium examples.
Moment = force x perpendicular distance from pivot.
Take moments (clockwise and anti-clockwise) about one of the supports.
Clockwise moment = anticlockwise moment
These equations should enable you to solve for the mass (weight) of the second student and find the rest of the information requested.
 
In other words torque is the force applied times its perpendicular distance from your chosen axis of rotation. Since it's not really rotating you can choose the rotational point anywhere convenient, like at the point where the person of unknown mass is located. If you sum all your torques about that point, since the acceleration is zero, that sum will be zero. do the same for just the forces and you have two equations.
 
Moment= 490 N * 2m = 980N on first support Moment on second support = Mass * 9.8 * 1m ?? how do i solve for mass of second student?
 
So from the point of the person of unknown mass, torque on closest support is M*9.8*1m, and the torque on the furthest support is M*9.8*9m?
 
  • #10
It is absolutely true... you can take any point you want and (I say 'take moments' !)
get expressions for the torques. There is zero resultant torque.
 
  • #11
88.2M + 9.8M + 9.8*120kg = Support1 + Support2?
 
  • #12
There are 3 unknowns...F1, F2 and the weight W of the second student.
So you will need 3 equations to get a solution. The usual thing is to take moments about each of the unknowns to eliminate it. Take moments about F1 will give you one equation, moments about F2 will give you a second equation and moments about W (the second student) will give you the third equation... Messy but should be able to solve these.
 
  • #13
wtf is a moment? I am so confused
 
  • #14
The problem cannot be solved without knowing the mass or weight of the 2nd student! Check the problem to see if it is given.
 
  • #15
jjd:
I am not sure how much you know about 'moments' (torque is more or less the same thing)
I started this question by drawing a diagram like this :
Draw a horizontal line to represent the 12m plank. At the mid-point draw a vertical down arrow attached to the plank and label it 1200N ( I am going to take g = 10 so that I can do calculations quickly in my head, you will need to use 9.81 to get a more exact answer).
At a point that looks like 2m on your plank line draw 2 supports... 1 at each end...2m from the end.
At each of these supports draw a vertical upwards arrow and label them F1 and F2 (I have F1 on the left) These are the forces at the supports on the plank.
On the left hand end of the plank draw a vertical down arrow labelled 500N... this is the first student
On the right hand end of the plank draw a vertical downward arrow what looks like 1m in from the end (1/2 way between end of plank and F2) label this W (this is the weight of the second student.
You should have a diagram with 3 down arrows (500N, 1200N and W Newtons)
and 2 up arrows F1 and F2.
Now you need to do that magic thing ... take moments.
Hope this helps.
 
  • #16
i already had that diagram drawn out, what i am having trouble with is taking these "moments" not sure how to do that
 
  • #17
I checked the problem, it is definitely not given
 
  • #18
I will show you how I started BUT I have a feeling that Phantomjay is correct although I do not know why?
Here goes on how to take moments...(bear in mind I am using g = 10)
Moments about the F1 support
Clockwise (1200x4) + (Wx9) = 4800 +9W
Anticlockwise (500x2) +(F2x8) = 1000 + 8F2
Clockwise = anticlockwise so...
4800 +9W = 1000 + 8F2
3800 = 8F2 - 9W...equ1
Moments about F2 gave me
9800 = 8F1 + W ...equ2
Moments about W gave me
11500 = 9F1 + F2 ...equ3
There is no resultant force so:
F1 + F2 = 1200 + 500 + W
F1 + F2 = 1700 + W ...equ4

I thought that with 4 equations relating 3 unknowns I would be able to solve this... BUT I have not been able to and I don't know why !
Your problem is not solved but at least I hope you can see a technique for solving this sort of problem
 
  • #19
moment i.e.torque
you can actually run 4 equations by putting your axis of rotation at each of the three unknowns and summing the torques and then using Newton's second law for linear motion also. If there is a missing piece of info it will show up in the solution of the four equations.

example. Put the axis of rotation at upper support 1 on the left

You then have -50kg (g)(2m) + 120kg(4m) - Fn'(8m) + m(person)(g)(9m) = 0

Repeat summing the torques with axis of rotation at Fn and again at m(person)(g) on the right)

Then sum the total forces on the system and try to solve from there.

Watch your signs. Designate counter clockwise torques negative and clockwise as positive.

I did get a solution but could have made a calculation error. I used the torques around the two upper supports to solve for each of them in terms of the the unknown person's weight. From there I put it back into the linear equation to get the unknown person's weight. Then just substituted to get the upper supports.

do you have the correct answer for it??
 
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  • #20
Phanthomjay... any input please
 
  • #21
netgypsy
I thought I had signs taken into account by calling moments clockwise and anticlockwise !
 
  • #22
I can only think that not all of these equations are independent but... I can't see it!
 
  • #23
jjd101 said:
I checked the problem, it is definitely not given

Then try as you might, you'll just be spinning wheels and going nowhere, because the problem cnnot be solved unless you know the 2nd student's weight. In the meantime, you should study up on moments and note that for systems in equilibrium, the sum of moments about any point (force times perpendicular distance) must equal zero (clockwise moments are plus and counterclockwise moments are minus), the sum of forces in the vertical direction must equal zero (up positive and down negative), and the sum of forces in the horizontal direction must equal zero (to the right as plus and to the left as minus).

In this problem, there's nothing going on in the x direction, so that latter equation in the x direction is useless here. That leaves you with the other 2 equations but with 3 unknowns (F1 reaction, F2 reaction, and 2nd student weight W), so you can't solve the problem without more information.
 
  • #24
This problem is way beyond my knowledge of physics so I am thinking maybe my professor made a mistake with the problem by not giving weight of second person? if anyone solves this let me know
 
  • #25
TECH You did - anticlockwise is fine with me. I was typing while you were answering.

I did solve it. Do you have the answer? I did the calculations in my head and am a bit rusty. Where IS my sliderule hahaha
 
  • #26
jjd101 said:
This problem is way beyond my knowledge of physics so I am thinking maybe my professor made a mistake with the problem by not giving weight of second person? if anyone solves this let me know
It is not solveable, so your prof apparently forgot to give you the student's weight. If we assume that the second student also has a mass of 50 kg, can you solve the problem now??
 
  • #27
No this is not beyond your knowledge of physics. It's just choose an axis of roatation, sum the torques, repeat three more times, sum the forces and solve the algebra II
 
  • #28
Phantomjay
I am confident about +/- moments.
I put them on the appropriate side of the = sign
I realize the x direction is of no consequence and I do not think it appears in any of these equations.
Can you see anything wrong in any of my 4 equations that come from the conditions for equilibrium.
Can you identify any equations that are not independent! That is the only way I can think there is something wrong
 
  • #29
netgypsy i do not have the correct answer for it, the homework is online so if i put in the answer it tells me if it is correct or not. Phanthom i think i could solve the problem then yes.
 
  • #30
Phantom I did solve it so again do you have the answer??jjd?
 
  • #31
OK let me finish solving for the two normal forces and plug them back in and double check my work
l8r
 
  • #32
netgypsy
You have solved it ! let us know... I am tearing my hair out... I feel it can be solved
 
  • #33
technician said:
Phantomjay
I am confident about +/- moments.
I put them on the appropriate side of the = sign
I realize the x direction is of no consequence and I do not think it appears in any of these equations.
Can you see anything wrong in any of my 4 equations that come from the conditions for equilibrium.
Can you identify any equations that are not independent! That is the only way I can think there is something wrong
The 2 applicable equations here are sum of Fy forces = 0 and sum of moments about any point =0. Only 2 equations are independent. If the weight of both students were given, you could sum vert forces=0 and sum moments about any point = 0, and get the answer for the 2 reaction forces.
Or you could sum moments about any 2 points to get the answer, but summing moments about another point, or even summing forces in the vert direction, does not give an independent equation, it just serves as a check on your work.
 
  • #34
Thank you phantom jay!
I felt there was something not quite right about the unknown W appearing in both the resultant force =0 equation and also in the sum of moments =0 but I could not see it so clearly.
Learn something every day !
 
  • #35
Phantom you are correct as I'm sure you already know. I ran it again and got an identity (it did help to get a calculator) so the best solution is to express the two force normals in terms of the unknown mass of the person.

Intuitively - as the mass of the person gets larger at the point 1 meter from the right side the normal force on the right will get larger, faster than the normal force on the left so as Phantom said, no single numerical answer is possible without one of the three variables as the information given does not restrict the weight of the person to a single value, (I'm not from Missouri but I don't take anyone's word for a solution until I run it) :-)

So you can select any mass for the person, solve the equations, select another mass, repeat, to show that there is not one exclusive mass that answers the question.

That was fun. don't you hate it when a problem is missing info. If it tells you when you get it right you could try some random masses and see if you can find the missing mass.
 
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  • #36
UPDATE: I just got an email from the teacher and the mass of the second student is 70kg, i calculated this with my understanding and got 1274N on right support and 1568N on left support if i am wrong please correct me
 
  • #37
Plug them into sum of forces = zero and sum of torques = zero and see if they work.
 
  • #38
the way i evaluated they even out. i got torque on first support from student one is 980 and torque on support two from student two is 656, both supports share weight of bar evenly which is 588 each
 
  • #39
No the supports do not necessarily share the weight evenly. You have to use force upward on the support on the left as a different variable from force upward on the support on the right. Your answers prove they are in fact different. It's two equations with two unknowns. Use an axis of rotation at one of the upward supports for your torque equation.
 
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  • #40
even though they are both placed in the same spot relevant to the bar?(2 feet from the edge) ?? how do i solve this then?
 
  • #41
it takes two equations. sum the forces and set equal to zero and sum the torques and set equal to zero.

Or Add the up forces and set them equal to the down forces and for your second equation add the clockwise torques and set them equal to the counterclockwise torques.
 
  • #42
Support 1 Force + Support 2 Force = Torque on support 1 + torque on support 2 + weight of beam? how do i deal with clockwise torques i am unfamiliar with this
 
  • #43
No do only forces for one equation - add the up forces and set them equal to the down forces. You'll have a variable for the left upward support force and another one for the right upward support force.

then do only torques for the second equation
 
  • #44
force of student one = 9.8*50 = 490 force of student 2 = 9.8*70 = 686 force of beam = 9.8*120= 1176 Total down force =2352 = Support 1 force + support 2 force?
 
  • #45
correct so far so good.

now you need to put the axis of rotation or your "pretend" point around which this plank will rotate at one of the two unknown upward forces and then add the clockwise torques and set them equal to the counter clockwise torques. Remember you have the equation listed torque = force times perpendicular distance from the force to the axis of rotation.
 
  • #46
so counter clockwise from first support, 490* 2
counter clockwise from second support 490* 10
clockwise from first support 686*9
clockwise from second support 686*1
these are not even though
 
  • #47
The first one on the left should be the guy at the end. He is two meters from the axis of rotation and he is counter clockwise. so 490N (2m) is correct for that one. The nextcounterclockwise is the upward force from the second support which is F right support (8m not 10meters) Your axis of rotation is at the left support which is 8m from the right support.

Clockwise you have the weight of the board downward times its distance from the left support. You don't have this one at all. and the last torque clockwise is from the guy on the right who is 1 meter from the right end so that one is the 686N(9m) so you are missing the clockwise torque caused by the weight of the board in the middle.

Once you find it, add the clockwise and set equal to the counterwise torques and solve for your only unknown. Then plug it back into the first equation you found
 
  • #48
clockwise torque of the board is 1176* 4m = 4704?
 
  • #49
Very good. So you see when you sum the clockwise and counterclockwise torques and set them equal to each other you have only one unknown, the upward force on the right side which produces a counterclockwise torque.

than from equation 1 you wrote summing just the forces, you can find your other upward force on the left side.
 
  • #50
clockwise = counter clockwise
clockwise = 4704+686*9 = 10875
counterclockwise = 490*2 + Fof2ndSupport*8
solving for this 2nd support is a force of 1237N therefore the force of the first support is 2352-1237 = 1115N ?
 

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