Are My Calculations on Ladder Equilibrium and Torque Correct?

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An extended object is in equilibrium when the net force and net torque acting on it are both zero. For a ladder placed against a smooth wall, the calculations show that the normal force from the wall is approximately 117.72 N and the force at the base is about 33.98 N. The minimum coefficient of friction required for stability is calculated to be 0.289. The discussion also emphasizes the importance of ensuring that both translational and rotational dynamics are balanced. The original poster seeks confirmation on the accuracy of these calculations.
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1) Give the conditions for the equilibrium of an extended object.
An extended object is at equilibrium when no force is acting on it.

2) A uniform ladder of length 6.0m and mass 12.0kg is placed against a smooth vertical wall. The ground is rough.

a) Calculate the force on the ladder from the wall.
vertical forces = 0 = N1 - W
so N1 = 117.72
horizontal forces = 0 = F - N2
Total Torque = 0
0 = F(0) + N1(0) + (12*9.81*cos60*(6/2)) - (N2sin60(6))
so, N2 = 33.98 Newtons

b) Find the minimum coefficient of friction of the ground.
F - 33.98 = 0 ---> F=33.98
friction > F1/N1
so, coefficient of friction=33.98/117.72 ----> 0.289

Are my answers correct? If not, what did I do wrong. Please help. Thanks! :smile:
 
Last edited:
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buffgilville said:
1) Give the conditions for the equilibrium of an extended object.
An extended object is at equilibrium when no force is acting on it.

2) A uniform ladder of length 6.0m and mass 12.0kg is placed against a smooth vertical wall. The ground is rough.

a) Calculate the force on the ladder from the wall.
vertical forces = 0 = N1 - W
so N1 = 117.72
horizontal forces = 0 = F - N2
Total Torque = 0
0 = F(0) + N1(0) + (12*9.81*cos60*(6/2)) - (N2sin60(6))
so, N2 = 33.98 Newtons

b) Find the minimum coefficient of friction of the ground.
F - 33.98 = 0 ---> F=33.98
friction > F1/N1
so, coefficient of friction=33.98/117.72 ----> 0.289

Are my answers correct? If not, what did I do wrong. Please help. Thanks! :smile:

I'm not going over your problem,but my guess is that:
An extended object is at equilibrium when the resulting momentum of all forces acting on it is nil and the result of all forces acting on it is nil.
It's placing a zero at the left of the dynamics' second law for the rotation movement and for the translation one.
 
Last edited:
What's nil ?
 
buffgilville said:
What's nil ?

ZERO,NAUGHT,NOTHING,ZIP,LOVE ?
Don't u like football??
 
Can someone help me check my work on the other problems, and tell me if they are correct?
 
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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