The system shown is in equilibrium. It is known that the moment at E is 1200ib-f
ID: 1837256 • Letter: T
Question
The system shown is in equilibrium. It is known that the moment at E is 1200ib-ft counterclockwise and that the weight of block (J) on the incline is 10001b. Find: W_a and W_b, tension coming off the pulleys at C, tension DG, force in bar GH and tension (T) attached to block J, f) normal force on the block, g) stiffness of spring 5, h) forces in each of the springs. Springs 1, 2, 3 and 4 have stiffness's of 40001b/jft, 30001b/ft. 2807.51b/ft, and 1422.51b/ft, respectively. Springs 1 and 5 are observed to deform 0.4' and 0.5', respectively.Explanation / Answer
Let the tension coming off the pulley at C is T1 lb and the tension DG be T2. lb The horizontal component of T2 is
T2cos angle Dg makes with horizontal=T2X (5/13).
The moment at E in anticlockwise direction is T1X10 lb-ft. and the moment at in clockwise direction is
T2X(5/13)x5.lb-ft
Thus net momentum at E in anti-clockwise direction is
T1X10-T2X(5/13)X5= 1200 lb-ft (i)
Forces on first four springs may calculated as
Spring-1: ( 0.4/12)X4000= 133.33 lb 1=t
Spring-2: (0.4/12)X3000= 100lb = t2
Spring-3: (0.4/12)X2807.5= 93.58lb= t3
Spring-4: (0.4/12)X 1422.5= 47.42 lb =t4
Component of the weight of the block j along inclined plane is 1000X sin 30*=500 lb
But T= t1+t2+t3+t4 +500= 874.33 =tension attached to the block j
We alo have t5= t1+t2+t3+t4= 374.33
where t5 is the tension in spring-5. it gives
If k5 is the stiffness of the spring-5 then we have
k5X (0.5/12)= 374.33
which gives k5= 8983.92 lb/ft lb/ft (stiffness of spring-5)
Angle bewtween GH and T is 30* and hence force in bar GH is T cos 30*= 874.33X( squareroot3/2)= 757.19 lb
Hence T2= 874.33-757.19= 115.15 lb
Substituting this value in eqn. (i) we have
T1X10=1200+ 115.15X(5/13)X5=1421.44lb-ft
Which gives T1= 142.144 lb
Here WA+WB =2400 lb
and WA-WB =T1= 142.144
which give
WA=1271.07 lb and WB=1128.93 lb