See picture top one The thin plate ABCD is held by the cable BH and by the links
ID: 1860076 • Letter: S
Question
See picture top one
The thin plate ABCD is held by the cable BH and by the links AE and DF. The mass oi the plate is 18[kg]. The length of each link is 150[mm] and the plate width is 600mm] and height of 300[mm]. Neglecting the mass of the links, determine, immediately after cable BH has been cut: The forced in links. Determine the acceleration of the plate. slender bar AB is 5[ft] long and is pivoted about point 0 l[ft] from the end B. e other end is pressed against a spring of constant k=1800[lb/in] until the spring isExplanation / Answer
f1 = FORCE IN LINK EA
F2 = force in link FD
T1 = tensio in BH
balancing horizental component of force :: F1*cos30 + F2*cos30 =T1*cos45
F1 + F2 = T1*2^0.5/3^0.5 = 0.81*T ..........(1)
Balancing Vertical component ::: F1*sin30 + F2*sin30 +T1*sin45 =18g
F1 + F2 + T1*0.414 = 18g ..........(2)
form (1) and (2) T1 = 143.4
balancing the torque about point A :: F2*cos30 *0.3 + 18*g*0.3 = T1*sin45 *0.6
F2 = 30 .5 N
form (1) F1 = 85.64 N
when string breaks T1 =0
a) force in links are F1 = 85.64 N
F2 = 30.5 N
b)
ax = horizental acceleration =[ F1*cos30 + F2*cos30]/18 = 5.58 m/s^2
ay = vertical acceleration = [18g-F1*sin30 + F2*sin30]/18 = 6.57 m/s^2
a = resultat acceleration = [ax^2 +ay^2]^0.5 = 8.6 m/s^2