Why Does Ladder Slipping Depend on the Ground's Surface?

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Ladder slipping is influenced by the friction between the ladder's base and the ground surface. A rough surface provides greater frictional force, helping to stabilize the ladder, while a smooth surface reduces this friction, making it easier for the ladder to slip. The forces acting on the ladder include gravitational force, normal force from the ground, and frictional forces at both the base and the top against the wall. When the ground is smooth, the frictional force at the base is insufficient to counteract the forces acting on the ladder, leading to instability. Understanding these dynamics is crucial for ensuring safety when using ladders.
brad sue
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Hi,
I need some explanation about a problem:

A heavy man stands on a ladder that is leaning against a rough-surfaced wall and rests on the ground. Dranw the free-body diagram for the system of the man and the ladder. If the ground is very smooth, the ladder may stast to slip. Why.

I need some explantions for the WHY question .

Thanks
 
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Explanation of WHAT? Do you know what a "free-body" diagram is? Can you determine the various forces acting here? What are the forces on the man? What are the forces on the ladder?
 
HallsofIvy said:
Explanation of WHAT? Do you know what a "free-body" diagram is? Can you determine the various forces acting here? What are the forces on the man? What are the forces on the ladder?

Yes, I found the forces on the man and on the ladder.
On the man there is only the gravity.

On the ladder, on the botton I have the reaction with the ground and a friction force directed toward the wall. On the top of the ladder, there are a reaction force coming out the wall and a vertical friction force.

I need explantion why the ladder starts to slip if the ground is very smooth.
 
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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