A ladder of mass M and length 2 m rests against a frictionless wall at an angle of 47 degrees to the horizontal. The coefficient of static friction between the ladder and the floor is 0.5. What is the maximum distance along the ladder a person of mass 13M can climb before the ladder starts to slip?
3 answers
i sitll have no idea how to do this. please help!
Draw the free-body diagram.
Pivot is on the floor.
Normal force of floor is pointing up(Nf).
Friction is in positive direction on the horizontal(f).
Normal of the wall is pointing in negative direction(Nw).
+ for counterclockwise, - for clockwise motion
remember that the θ in rFsinθ is the angle between r and F so draw the ladder cutting into the wall so you can draw the angle between the ladder and Nw.
ΣFy=0; Nf-13Mg-Mg=0 ∴ Nf=14Mg
ΣFx=0; -Nw+f=0 ∴ Nw=f
Nw=f=μNf=[μ14Mg] (you'll need this later)
Στ=0; Nf(0)+[Nw]sin133°+(d-(2/2)13Mgsin137°-(2/2)Mgsin137°=0
replace Nw and solve for d.
Pivot is on the floor.
Normal force of floor is pointing up(Nf).
Friction is in positive direction on the horizontal(f).
Normal of the wall is pointing in negative direction(Nw).
+ for counterclockwise, - for clockwise motion
remember that the θ in rFsinθ is the angle between r and F so draw the ladder cutting into the wall so you can draw the angle between the ladder and Nw.
ΣFy=0; Nf-13Mg-Mg=0 ∴ Nf=14Mg
ΣFx=0; -Nw+f=0 ∴ Nw=f
Nw=f=μNf=[μ14Mg] (you'll need this later)
Στ=0; Nf(0)+[Nw]sin133°+(d-(2/2)13Mgsin137°-(2/2)Mgsin137°=0
replace Nw and solve for d.
oops my bad, the weight of the person is in clockwise direction so it's negative
Στ=0; Nf(0)+[Nw]sin133°-(d-(2/2))13Mgsin137°-(2/2)Mgsin137°=0
Στ=0; Nf(0)+[Nw]sin133°-(d-(2/2))13Mgsin137°-(2/2)Mgsin137°=0
1 answer