Why is the reaction force angled upwards on a flat drawbridge?

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The reaction force on a flat drawbridge is angled upwards due to the combination of horizontal and vertical forces acting on it. A horizontal force alone would not prevent the bridge from falling, as it requires a vertical component to support its weight. If the wall were frictionless, the bridge would lack the necessary vertical support and would fall. The bridge is held up by a normal force from the wall, which is a result of compression, along with static friction if applicable. Thus, the angled reaction force is essential for maintaining the bridge's stability.
danago
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Hi. Not directly a specific homework question, but it will help with some of my questions.

When the bridge is flat, why is the reaction force from the wall on the bridge angled upwards? Why is it not horizontal, along the length of the bridge?

Sorry if i was a little vague.

Thanks,
Dan.
 
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If the force were only horizontal, how would that keep it from falling? I'm also sorry to be vague.
 
Hmm ok. If it was a frictionless wall, would it be possible for an angled reaction force to exist?
 
I don't understand the situation. Are you saying that the drawbridge is held up by friction? How is the bridge attached to the wall? What pulls the bridge up and down?
 
Nearest I can come to having it make sense is that the bridge is being compressed against the wall. So there two forces. There is the horizontal wall reaction force to the normal force of compression. And there would be a vertical static frictional force holding the thing up. So the combination of these two forces can be 'angled upwards'. If it's a frictionless wall, the bridge will fall.
 
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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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