# Boom vector problem

• Jul 21st 2008, 01:34 PM
sarnekat
Boom vector problem
Hi can anybody help me in this question that i have???

A boom OB carries a load F of manetude 500N and is supported by cables BC as shown in the diagram where the dimensions of the system are given. Determine the tensions in the cables so that equilibrium is maintained and the resultant force at piont B as shown in drawing attached.(Headbang)
• Jul 21st 2008, 06:53 PM
ticbol
Quote:

Originally Posted by sarnekat
Hi can anybody help me in this question that i have???

A boom OB carries a load F of manetude 500N and is supported by cables BC as shown in the diagram where the dimensions of the system are given. Determine the tensions in the cables so that equilibrium is maintained and the resultant force at piont B as shown in drawing attached.(Headbang)

In addition to the tension T in cable BC, you want to know also the tension S in cable BD, I think.
In your sketch, I understand that the foot of the boom, point O, is at (0,0,0) of the (x,y,z) axes.
Also, the points B,C,D are all in the same z = 75cm plane....and this plane is horizontal.
And, point B is at (100cm, 0 , 75cm).

In the vertical (x,z) plane....for the compression U in the boom:
U is in the hypotenuse of the right triangle whose legs are 75cm vertical and 100cm horizontal.

At point B,
Break the U into its vertcal and horizontal components Uz and Ux respectively.
For equilibrium in the z-direction,
Uz = F = 500N.

By proportion, Uz / 75cm = Ux / 100cm
So, Ux = (100 / 75)Uz = (100 / 75)(500N) = 666.66667 N -------------------------------------**
This Ux will be supported by the combined tensions S and T in the two cables.

----------------------------------
In the horizontal z = 75cm plane:

For the tension T in the BC cable,
T is in the hypotenuse of the right triangle whose legs are 50cm and 100cm....so, the hypotenuse is sqrt(50^2 +100^2) =
111.8034 cm.
Break the T into its Tx and Ty components,
By proportion,
T / 111.8034 = Tx / 100 .......Tx = (100 / 111.8034)T = (0.89443)T
T / 111.8034 = Ty / 50 ......Ty = (50 / 111.8034)T = (0.44721)T

For the tension S in the BD cable,
S is in the hypotenuse of the right triangle whose legs are 75cm and 100cm....so, the hypotenuse is sqrt(75^2 +100^2) = 125cm.
Break the S into its Sx and Sy components,
By proportion,
S / 125 = Sx / 100 .......Sx = (100 / 125)S = (0.8)S
S / 125 = Sy / 75 ......Sy = (75 / 125)S = (0.6)S

For equilibrium in the y-direction at point B,
Ty = Sy
(0.44721)T = (0.6)S
T = (0.6 / 0.44721)S = (1.34165)S ------------------**

For eqilibrium in the x-direction,
Ux = Tx +Sx
666.66667 N = (0.89443)T +((0.8)S
Substituting (1.34165)S for T,
666.66667 = (0.89443)(1.34165 S) +(0.8)S
S = 666.6667 / 2.0