The system shown is in equilibrium. It is known that the moment at E is 1200ib-f

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

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