4) A 30 ft ladder weighing 80 lbs rests against a slippery (frictionless) wal Th

Question

4) A 30 ft ladder weighing 80 lbs rests against a slippery (frictionless) wal The foot of the ladder is 20 feet from the wall. The coefficient of static friction between the ground and the bottom of the ladder is 0.5, and the coefficient of sliding friction is 0.4. A person weighing 160 lbs climbs up the ladder. (a) Calculate the horizontal contact force Fx of the wall on the ladder, as well as the vertical contact force Fy and the frictional force f on the ladder by the ground required to hold the person when they are half way up. (b) How far up the ladder can the person climb before the ladder will be in danger of sliding out?

Explanation / Answer

You have already drqawn the FBD

Let at x-distance from the bottom of slide i will slide out.

Resolving forces in horizontal direction.

Fx-f=0

Fx=f=0.5(80+160)*9.8=1176 lbs. ft/s^2

b)Resolving forces in vertical direction.

Fy=(m+M)g=(80+160)*9.8=2352 lbs. ft/s^2

Frictional force=f=0.5(80+160)*9.8=1176 lbs. ft/s^2

b)Calculating torgue at bottom point when the peson is at x-distance fron ladder.

mg*(L/2)cos theta + Mg*x cos theta- Fx*L sintheta=0

x=(Fx*L sintheta-mg*(L/2)cos theta )/(Mg* cos theta )

x=(1176*30*sin48.19 - 80*9.8*15 cos48.19)/(160*9.8*cos48.19)=16.58 ft

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