A piece of metal is held in place as shown At point A the pi

A piece of metal is held in place as shown. At point A the piece is held in a threat bearing (which does not allow translation along the y-axis). A force is applied at point D as shown. A cable at B attaches the place to the wall at C. The tension in the cable is 10 N and the weight of the piece can be assumed to be negligible compared to this. Determine the loads (forces and/or moments) acting at A and the unknown applied force at D. Express your answers in vector notation. Describe in words why you represented the thrust bearing at A as you did in your fee body diagram of the system.

Solution

1) here coordinate of point A,B,C andD are A(0,0,0),B(3,6,0),C(3,0,2) and for D(1.5,6,0)

wheret=10n and force at D=Fo N

2)here unit vector along CD are

rcd=(x2+x1)i+(y2-y1)j+(z2-z1)k

rcd=-1.5i+6j-2k

magnitude of rcd=lrcdl=6.5

hence unit vector cd is =-.2307i+.923j-.3076k

4) tension vector is

T\'=T*rcd unit vector

T\'=-2.307i+9.23j-3.076k

5) force acting at D is

Fd=-Foj

6) now taking moment aroun poinA we get

Ma=T\'*rab+Fd*rad

on taking vector cross product we get

T\'*rab=18.456i-9.228j-41.532k

and Fd*rad=1.5Fok

on summation we get

Ma=18.456i-9.228j+(-41.532+1.5Fo)

where at A,we have thrust bearing moment around x,y is possible but around z is zero

Mz=0

Ma=Mxi+Myj+Mzk

on equating we get

Mx=18.456 Nft

My=-9.228 Nft

Mz=-41.532+1.5Fo=0

Fo=27.68 N

7) taking vector sum at equillibrium we get

T\'+Fd+A\'=0

we get reaction at A as

A\'=2.307i+(9.23-27.68)j+3.076k

A\'=2.307i+18.45j+3.076k

here A\'=Axi+Ayj+Azk on equating we get

Ax=2.307 N

Ay=18.45 N

Az=3.076 N