The boom ABC below is connected to the wall at point C by a

The boom ABC below is connected to the wall at point C by a ball-and-socket joint and by two cables BE and AD. a. Draw a FBD of the boom ABC b. Calculate the reaction forces acting on the boom at point C and the tensions in cables AD and BE.

Solution

solution:

1)here boom ABC is subjected force f at A and tension Tbe and Tad at b and A point respectively with support reaction at point C with certain moment.

2)here coordinate of points are

C(0,0,0),A(2.4,0,0),B(1.8,0,0),D(0..3.1.2),E(0,.6,-.9)

3)here tension along BE and AD are given by

Tbe=T1.Rbe^=-.8571Ti+.2857T1j-.4285T1k

Tad=T2.Rad^=-.888T2i+.111T2j+.444Tk

4)here moment around point D and E are

Md=0

Fa=0i-800j-500k

Rad X Fa+Tbe X Red+R X Rcd=0

so finally we get that for i,j,k part as

Md=[810-.4714T1+.3Rz-1.2Ry]i+[-1200+1.7999T1+1.2Rx]j+[1920+.2571T1-.3Rx]k=0

here for j and k part solve linearly gives us

Rx=4436.50 N

T1=-2291.12 N

and equationas

1.2Ry-.3Rz=1890.03

5)for taking moment around point E

Me=0

Rae X Fa+Tad X Rde+R X Rce=0

here on solving we get that for j and k part as

T2=-2781.56 N

Rx=4435.01 N

and relation as

.9Ry+.6Rz=2038.88

5)in this way above relation for Ry and Rz gives

1.2Ry-.3Rz=1890.03

.9Ry+.6Rz=2038.88

on solving we get that

Ry=1763.31 N

Rz=753.16 N

6)hence tension in cable are

T1=-2291.12 N

T2=-2781.56 N

Rx=-4435.01 N

Ry=1763.31 N

Rz=753.16 N