A 32.8kg, uniform ladder of length L hinged to a horizontal platform at Point P1
ID: 2252738 • Letter: A
Question
A 32.8kg, uniform ladder of length L hinged to a horizontal platform at Point P1 and anchored with a a steel cable of the same length as the ladder attached at the ladder's midpoint. The ladder makes an angle of 50 degrees with the horizontal.
A. Calculate the magnitude of the tension in the cable when a 75.5kg person is standing three-quarters of the way up the ladder.
Answer given is 2000N.
B. Calculate the components of the force in the hinge when an 75.5kg person is standing three-quarters of the way up the ladder (include direction)
Answer given is Fx= 1840N, Fy= 1830N
Explanation / Answer
let theta be the angle made by the cable with horizontal
sin50 = y/(L/2)
sin(theta) = y/L
sin50/sintheta) = 2
sintheta = (sin50)/2
theta = 22.5 degrees
M= 32.8 kg
m = 75.5 kg
net torque = 0
-T*cos22.5*(L/2)*sin50 + T*sin22.5*(L/2)*cos50 + m*g*(3/4)*L*cos50 + M*g*(L/2)*cos50
T*cos22.5*(L/2)*sin50 - T*sin22.5*(L/2)*cos50 = m*g*(3/4)*L*cos50 + M*g*(L/2)*cos50
(T/2)*[sin50*cos22.5 - cos50*sin22.5 ] = m*g*(3/4)*L*cos50 + M*g*(L/2)*cos50
T/2*sin(50-22.5) = m*g*(3/4)*L*cos50 + M*g*(L/2)*cos50
T/2*sin(50-22.5) = (75.5*9.8*(3/4)*cos50) + ( 32.8*(1/2)*9.8*cos50)
T = 1994.5 N = 2000N
along x axis
Tx + Fx = 0
Fx = Tx = T*cos22.5 = 1840 N
along y axis
Fy - Ty - (m+M)*g = 0
Fy = 1827.8 = 1830 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.