http://s2.postimg.org/t6cvvfw95/10_2.png The figure shows a 29.9-kg, uniform lad
ID: 2262302 • Letter: H
Question
http://s2.postimg.org/t6cvvfw95/10_2.png
The figure shows a 29.9-kg, uniform ladder of length L hinged to a horizontal platform at point P1 and anchored with a steel cable attached at the ladder's midpoint. The angle, ?, between the ladder and the floor is 50.0
The figure shows a 29.9-kg, uniform ladder of length L hinged to a horizontal platform at point P1 and anchored with a steel cable attached at the ladder's midpoint. The angle, ?, between the ladder and the floor is 50.0 degree, and the angle, ?, between the rope and the floor is 30.0 degree. Calculate the tension in the cable when a 73.9-kg person is standing three-quarters of the way up the ladder. Calculate the horizontal force component in the hinge when a 73.9-kg person is standing three-quarters of the way up the ladder. Calculate the vertical force component in the hinge when a 73.9-kg person is standing three-quarters of the way up the ladder.Explanation / Answer
Tension in the cable is C.
Horizontal force in the hinge is P1x.
Vertical force in the hinge is P1y.
The mass of the ladder is m1.
The mass of the person is m2.
Start off with summing your forces and moments, and equating them to zero:
summation Fx = 0 = C*cos(theta) - P1x
summation Fy = 0 = P1y - m1*g - m2*g - C*sin(theta)
summation M about ladder CG = 0 = m2*g*L/4*cos(alpha) + P1y*L/2*cos(alpha) - P1x*L/2*sin(alpha)
Rearrange the moment summation about the ladder CG:
0 = m2*g*L/4*cos(alpha) + P1y*L/2*cos(alpha) - P1x*L/2*sin(alpha)
P1x*L/2*sin(alpha) = m2*g*L/4*cos(alpha) + P1y*L/2*cos(alpha)
P1x/2*sin(alpha) = m2*g/4*cos(alpha) + P1y/2*cos(alpha) (canceled out L)
P1x = m2*g/2*cot(alpha) + P1y*cot(alpha)
substitute this equation into the summation of forces in the x direction and solve for P1y
0 = C*cos(theta) - P1x
0 = C*cos(theta) - m2*g/2*cot(alpha) - P1y*cot(alpha)
P1y*cot(alpha) = -m2*g/2*cot(alpha) + C*cos(theta)
P1y = -m2*g/2*cot(alpha)*tan(alpha) + C*cos(theta)*tan(alpha) = -m2*g/2 + C*cos(theta)*tan(alpha)
then plug this into the summation of forces in the y direction and solve for C
0 = P1y - m1*g - m2*g - C*sin(theta)
0 = -m2*g/2 + C*cos(theta)*tan(alpha) - m1*g - m2*g - C*sin(theta)
-C*cos(theta)*tan(alpha) + C*sin(theta) = -m2*g/2 - m1*g - m2*g
C*(sin(theta) - cos(theta)*tan(alpha)) = g*(-m2/2-m1-m2) = g*(-3*m2/2-m1)
Rather than solve for C at this point, I'm going to plug in some numbers:
C*(sin(30)-cos(30)*tan(50)) = 9.8*(-3/2*(73.9) - 29.9)
-0.532*C = -1379.35
C = 2592.76 N........................................(a)
Now use the summation of x forces to get P1x
0 = C*cos(theta) - P1x
P1x = C*cos(theta) = 2592.76*cos(30) = 2245.33 N............................(b)
and use the equation we plugged into the moment summation to get P1y
P1y = -m2*g/2 + C*cos(theta)*tan(alpha) = -73.9*9.8/2 + 2592.76*cos(30)*tan(50) = 2313.77 N ...............(c)
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