4) A 30 ft ladder weighing 80 bs rests against a slippery (frictiones) wall The

Question

4) A 30 ft ladder weighing 80 bs rests against a slippery (frictiones) wall 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 be in danger of sliding out?

Explanation / Answer

cos (theta ) = 20/30

theta = 48 degrees

Fx*L*sin48 - Wladder*L/2*cos48 - Wperson*L/2*cos48 = 0

Fx*sin48 - Wladder*1/2*cos48 - Wperson*1/2*cos48 = 0

Fx *sin48 - 80*1/2*cos48 - 1601/2*cos48 = 0

Fx = 756.8 lb

Fy = Wlader + Wperson = 240 lb

frictional force f =Fx = 756.8 lb

===========================================

Fx*L*sin48 - Wladder*L/2*cos48 - Wperson*l*cos48 = 0

f = us*Fy

Fx = f = us*Fy = 0.5*240 = 120 lb

Fx*30*sin48 - Wladder*30/2*cos48 - Wperson*l*cos48 = 0

120*30*sin48 - 80*30/2*cos48 - 160*l*cos48 = 0

l = 17.4 ft.........................

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.