A 15.5 m uniform ladder weighing 500 N rests against a frictionless wall. The la

Question

A 15.5 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 62.0 angle with the horizontal. A 15.5 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 62.0° angle with the horizontal (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 820 N firefighter is 4.20 m from the bottom Magnitude of the horizontal force Direction away from the wall towards the wall Magnitude of the vertical force Direction down up (b) If the ladder is just on the verge of slipping when the firefighter is 9.10 m up, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

As the ladder is in equilibrium then the net force acting on the ladder must be zero

Now let Fx be the force exerted by the wall on the ladder

Now applying the net torque about the bottom =0

500*(15.5/2)sin(28)+820*4.20*sin(28)-Fx*15.5*sin(62) =0

Fx*15.5*sin(62) =500*(15.5/2)sin(28)+820*4.20*sin(28)

Fx =[500*(15.5/2)sin(28)+820*4.20*sin(28)/15.5*sin(62)] =[1819.202+1616.860]/13.685=251.082N

And now Let Fbx and Fby are the forces exerted by the florr on the ladder

Applying Fnetx =0

Fbx-Fx =0

Fbx =Fx =251.082N

Now applying for Fnety =0

Fby-500-820=0

Fby =500+820=1320N

b)

Now about the lowest point

= 0

820 * 9.10 cos 62 + 500 * (15.5/2)*cos 62 - nw * 15.5 sin 62 = 0

3503.196+1819.202 = 13.685 nw

nw = 388.922 N

So, f = nw = 388.922 N

      fmax = f

      s ng = 388.922

      s = 388.922 / 1320

         = 0.2946

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