These... The 60 lb bar is supported as shown. Surface B is frictionless. Where m
ID: 1861622 • Letter: T
Question
These...
The 60 lb bar is supported as shown. Surface B is frictionless. Where must the 10 lb weight be placed to maintain equilibrium? Ans: 3.35 ft A 90 kg man is climbing a 5 m ladder that has a mass of 20kg. what are a) the reactions on the ladder when x = 1.5m and b) the smallest value for x that will make the ladder fall? (a)RA = 647 N, RB = 858 N, (b) 2.61 m The mechanism shown is designed to maintain tension on the wire shown during processing. If the device has a mass of 0.4 kg that can be idealized as being at point G and the spring force is 14 N, what is the tension in the wire and the magnitude of the force at A? T= 13.7 N,RA = 27.4 NExplanation / Answer
A.
let N be the normal force act on the wedge and T be the tension in wire
N act at an angle of 30 with horizontal
equilibrium condition
sum of net force in x direction = 0
T cos 35 = N cos 30
N = 0.94 T
sum of net force in y direction = 0
T sin 35 + N sin 30 = 60 + 10
on solving
T = 67.3 lb
net moment about point where Nacts = 0
(T sin 35 * 4ft )-(60 lb * 2 ft )- (10 lb * x ft) =0
put all value we get
x =3.35 ft
c.
net moment about A = 0
(T * 20mm )+(14 N * 120 mm ) -(T * 120mm) - ( 0.4g * 80mm ) =0
T= 13.7 N
LET FORCE ON A BE R
lr
let at A in -x direction force is Rx and in -y direction force is Ry such that
R=( Rx^2 + Ry^2 )^1/2
in y direction
13.7 + 14 = 0.4g + Ry
Ry =23.776 N
Rx = T = 13.7 N
R=( 23.776^2 + 13.7^2 )^1/2
R=27.4N
B.
FIGURE IS NOT CLEAR
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