A 16.0-m uniform ladder weighing 490 N rests against a rictionless wall. The lad

Question

A 16.0-m uniform ladder weighing 490 N rests against a rictionless wall. The ladder makes a 57.0Degree angle with the horizontal.Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 840-N firefighter has climbed 3.80 m along the ladder from the bottom. Horizontal Force magnitude N direction Vertical Force magnitude N direction If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

for the magnitude of the force of the wall, we will use summation of moment equals zeroassume positive as clockwise

let R be the magnitude of the force of the wall
M = 840(3.8*cos57) + 490(8cos57) - R(16sin57) = 0

=>0= 1738.49 + 2134.98 -R*13.41

then
R = 288.84N directed to the left
towards the wall

for vertical component, we have,

weight of ladder and man acting downwwards, hence reaction force,

Gy = 490 + 840 = 1330 N directed upwards

b) now similarly we find the vertical force , which will remain unchanged , that is 1330 N

now , at 9.4 m ,the horizontal reaction will be,

M = 840(9.8*cos57) + 490(8cos57) - R(16sin57) = 0

=>4483.46+2134.98= R(16sin57)

=>R=493.22 N

for the coefficient of static friction,

f = uN
493.22 N N = u(1330)
u = 0.370

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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