A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The la
ID: 1458454 • Letter: A
Question
A 14.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 63.0° angle with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.30 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 wallExplanation / Answer
(a)
Since the ladder is uniform, its center of gravity is right in the middle, 7.0 m from either end.
The first condition for equilibrium gives:
nwall - Fx = 0
Fy - 520 N - 850 N = 0
Fy = 1370 N Upwards
The second condition for equilibrium, choosing the base of the ladder as the axis of rotation, gives:
(520 N)(7.0 m)cos63° + (850 N)(4.30m)cos63° - nwall*(14.5 m)sin63° = 0
nwall = [(490 N)(7.25m)cos57° + (820 N)(4.2m)cos57°]/(14.5 m)sin57°
nwall = 256.34 N (Towards the wall.)
Horzontal Force , Fx = 256.34 N , Towards the wall.
Vertical Force , Fy = 1370 N , Upwards.
(b)
Torque = 850 * 8.9 * cos 63 = 7832 * cos 63
Total counter clockwise torque = (520)*(7.0)*cos63° + 850 * 8.9 * cos 63
To determine the horizontal force, set this equal to the clockwise torque
F * 14 * sin 63 = (520)*(7.0)*cos63° + 850 * 8.9 * cos 63
F = 407.8 N
This is the magnitude to the friction force.
Ff = * N
N is the total weight
Ff = * 1370
* 1370 = 407.8
= 0.297
Coefficient of static friction between ladder and ground, u = 0.297
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.