A 14.0-m uniform ladder weighing 510 N rests against a frictionless wall. The la

Q: A 14.0-m uniform ladder weighing 510 N rests against a frictionless wall. The la

a) here Ng = 820 + 510 = 1330 N up the wall then torque(wall) = torque(firefighter) + torque(ladder) Nw * cos28.5 * 14 = 820 * 4.1 * sin28.5 + 510 * 7 * sin28.5 Nw = 268.85 N to the left Ff = 268.85 N to the right b) cos28.5 * Nw * 14 = 820 * 8.9 * sin28.5 + 510 * 7 * sin28.5 Nw = 421.5 N then Nw = u * Ng 1330 * u= 421.5 u = 0.316

ID: 1551706 • Letter: A

Question

A 14.0-m uniform ladder weighing 510 N rests against a frictionless wall. The ladder makes a 57.0° angle with the horizontal.

(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 820-N firefighter has climbed 4.10 m along the ladder from the bottom.
Horizontal Force

Vertical Force

(b) If the ladder is just on the verge of slipping when the firefighter is 8.90 m from the bottom, what is the coefficient of static friction between ladder and ground?

magnitude N direction ---Select--- towards the wall away from the wall k -XX"x-)x-* )#7 *

Explanation / Answer

here

Ng = 820 + 510 = 1330 N up the wall

then

torque(wall) = torque(firefighter) + torque(ladder)

Nw * cos28.5 * 14 = 820 * 4.1 * sin28.5 + 510 * 7 * sin28.5

Nw = 268.85 N to the left

Ff = 268.85 N to the right

cos28.5 * Nw * 14 = 820 * 8.9 * sin28.5 + 510 * 7 * sin28.5

Nw = 421.5 N

then

Nw = u * Ng

1330 * u= 421.5

u = 0.316

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A 14.0-m uniform ladder weighing 510 N rests against a frictionless wall. The la

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