Looking for understanding of very basic force questions

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To determine the forces on two pylons from a steel beam, one must consider static equilibrium, where the sum of vertical forces and moments equals zero. Given a 700 N beam, the force distribution can be calculated by taking moments around one pylon and applying the principle that the upward forces must equal the downward weight. The calculations yield forces of 300 N on Pylon A and 400 N on Pylon B. This approach ensures that both translational and rotational motions are accounted for, confirming the static nature of the system. Understanding these principles is essential for solving similar problems in static mechanics.
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1. Can someone explain how to work out the force on 2 pylons from a steel beam laid between them.
Beam 3mtrs
Pylon A 30cm from end Pylon B 60cm from other end
Beam has 700 N weight



I have the answers of A 300N and B 400N but not sure how to get there.


Sorry but its over 30yrs since I did this and my daughter is just starting to learn.
Any help appreciated
 
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Since the body in static, there is no rotational or translational motion.
Take moment at any point, CCW moment equal to CW moment.
For traslational motion, upward force equal to downward force.
 
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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