Question

A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the coefficient of static friction between the floor and the ladder is 0.28. what distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?

Answer 1

First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

Answer 2

Answer:

d=1.18m

Explanation:

A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the coefficient of static friction between the floor and the ladder is 0.28. what distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?

frictional force =uFn

u=coefficient of static friction

Fn=normal force

assuming forces in the y direction, we ave

EFy=0

Efy=Fn-Fg-Fp

Fn=Fg+Fp

Fn=120N+600N

Fn=720N

we recall tat Frctional force=uFn

0.28*720N

201.6N

summing te forces in te x direction

EFx=Fr-fWall

Fr=frictional force

Fwall=force in the wall

Fr=Fwall

Fwall=201.6N

the angle te ladder makes wit round will be

cos$\alpha$=3/5

$\alpha$=53.13

Next, we can turn to calculating the net torque

Choose the pivot point at the bottom of the ladder

(this choice eliminates FN and Fr).

rgrav =1/2*5=2.5

rwall=5m

rman=d

twall== r( wall)(Fwall)sinalpa

5*201.6sin53.13

twall=+806.39Nm

τ grav = r( grav )(Fgrav )sin53.13

τ grav=2.5*120*)sin53.13

-239.99nm

τ person = r( person )(Fperson )sin53.13

d*600Nsn53.13

-479.99d

∑τ = τ wall + τ grav + τ person

= 0

summation of te torque

te negative sin is for troques in te clockwise direction

+806.39Nm-239.99nm-479.99d=0

566.4=479.99

d=1.18m