What Are the Forces on a Supported Beam?

[netgypsy]

OK let me finish solving for the two normal forces and plug them back in and double check my work
l8r

 

[technician]

netgypsy
You have solved it ! let us know... I am tearing my hair out... I feel it can be solved

 

[PhanthomJay]

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.

 

[technician]

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 !

 

[netgypsy]

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.

 

[jjd101]

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

 

[netgypsy]

Plug them into sum of forces = zero and sum of torques = zero and see if they work.

 

[jjd101]

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

 

[netgypsy]

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.

 

[jjd101]

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?

 

[netgypsy]

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.

 

[jjd101]

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

 

[netgypsy]

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

 

[jjd101]

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?

 

[netgypsy]

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.

 

[jjd101]

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

 

[netgypsy]

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

 

[jjd101]

clockwise torque of the board is 1176* 4m = 4704?

 

[netgypsy]

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.

 

[jjd101]

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 ?

 

[netgypsy]

You're doing it correctly. I haven't checked your math. Good job.

 

[jjd101]

Thank you!

 

[netgypsy]

no problem (hahaha)