Beam calculation with 3 unknowns

AI Thread Summary
A 5m metal beam supported by two points is analyzed to find reaction forces and additional weights placed to prevent tipping, with a total downward force of 931N from the beam and a man standing 0.5m from one end. The equations used include the sum of forces and moments, leading to two main equations involving three unknowns: the reaction forces (F1 and F2) and the additional weight (W) at support B. A suggestion is made to assume the weight at B is the minimum required to prevent tipping, which simplifies the calculations. There is a clarification regarding the beam's length and the man's position, confirming the beam is indeed 5m long. The discussion emphasizes the importance of correctly interpreting the problem to find a solution.
willowdanny
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Homework Statement


A 5m metal beam is supported on 2 supports. The beam is 30kg and the man is 65kg. The man stands at A, to prevent the beam tipping extra weights have been placed at B. Find reaction forces and the weights that have been placed at B. Take acceleration due to gravity to be 9.8ms
moment.png

Homework Equations


Force * Distance
Sum of Clockwise = Sum of Anti-Clockwise
Sum of Downward Forces = Sum of Upward Forces

The Attempt at a Solution


Decorator - 637N
Beam - 294N
F1 being the first support reaction force
F2 being the second reaction force
W being the extra weights added

F1 + F2 = 931 + W (Sum of up and down)

3F1 + F2 = 3601.5 (sum of clockwise/anticlockwise)

I have attempted so many different ways to figure this out, but I have no idea how to get around having 3 unknowns at a time. When I take a moment around one of the supports, I have the reaction force the other plus the weight being placed at B. I'm not sure if I'm missing something obvious or not.
Thanks in advance for literally any help
 

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Welcome to PF!

Maybe you are supposed to assume that the weight at B is the minimum weight that will prevent the beam from tipping.

I don't quite agree with your number 3601.5 in the torque equation. EDIT: Nevermind, I see that A is not at the end of the beam. The number 3601.5 looks good.
 
Last edited:
willowdanny said:

Homework Statement


A 5m metal beam is supported on 2 supports. The beam is 30kg and the man is 65kg. The man stands at A, to prevent the beam tipping extra weights have been placed at B. Find reaction forces and the weights that have been placed at B. Take acceleration due to gravity to be 9.8ms
View attachment 226011

Homework Equations


Force * Distance
Sum of Clockwise = Sum of Anti-Clockwise
Sum of Downward Forces = Sum of Upward Forces

The Attempt at a Solution


Decorator - 637N
Beam - 294N
F1 being the first support reaction force
F2 being the second reaction force
W being the extra weights added

F1 + F2 = 931 + W (Sum of up and down)

3F1 + F2 = 3601.5 (sum of clockwise/anticlockwise)

I have attempted so many different ways to figure this out, but I have no idea how to get around having 3 unknowns at a time. When I take a moment around one of the supports, I have the reaction force the other plus the weight being placed at B. I'm not sure if I'm missing something obvious or not.
Thanks in advance for literally any help

You say the beam is 5m in length, but the segments shown add up to only 4.5m. Which figure is erroneous?
 
Ray Vickson said:
You say the beam is 5m in length, but the segments shown add up to only 4.5m. Which figure is erroneous?
I overlooked that A is not at the end of the beam. So, the beam can be 5 m long.
 
Ray Vickson said:
You say the beam is 5m in length, but the segments shown add up to only 4.5m. Which figure is erroneous?
The beam is 5m in length, the man is standing 0.5m away from the end of the beam.
 
willowdanny said:
The beam is 5m in length, the man is standing 0.5m away from the end of the beam.
Have you tried assuming the added weight is the minimum, as TSny suggests? That is also how I read the question. Hint: it allows you to write down one reaction force immediately.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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