I have literally been working on this problem for HOURS! I can't get an answer that works. Please someone walk me through this.
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If the tension developed in each of the cables cannot exceed 340 lb, determine the largest weight of the crate that can be supported. Also, what is the force developed along strut AD?
Here is the picture: http://session.masteringengineering.com/problemAss...
Update:Here is the link if the above one didn't work:
http://imreglsu43.tumblr.com/post/31321261837
Verified answer
The point of interest is A. Begin with setting up the free-body diagram of A, with the four forces acting at A. There are no moments to calculate, as all the forces pass through A.
The four forces are as follows:
Weight of the Box, down. (Let's call this W)
Tension of AB, toward B. (We'll call it Tb)
Tension of AC, toward C. (We'll call this Tc)
"Tension" from AD, from D. (And we'll call this D)
To begin with, calculate the unit vectors of AB, AC, and AD. The unit vector for W is <0,0,-1>. The unit vectors for the others are, as previously ordered, <-4/13,-12/13,3/13>, <2/7,-6/7,3/7>, and <0,12/13,5/13>.
Breaking the force vectors into their components, we are left with the following equations:
(1) Fx = 0 = -4/13*Tb + 2/7*Tc
(2) Fy = 0 = -12/13*Tb + -6/7*Tc + 12/13*D
(3) Fz = 0 = 3/13*Tb + 3/7*Tc + 5/13*D - W
From (1), we can solve for Tb in terms of Tc, such that Tb = 13/14*Tc
From (2), we can substitute our solution from (1) into Tb and then solve D in terms of Tc, D = 13/7*Tc
Then from (3), we can substitute (1) and (2) for Tb and D and put W in terms of Tc, W = 19/14*Tc.
From (1), we can see that Tb will be less than Tc, so Tc shall be equal to 340 lbs.
W thus shall equal 460 lbs
D will equal 630 lbs
Set up equations that sum the moments about the vertical axis and horizontal axis passing through point D. Designate the weight of the crate by W. Manipulate your equations to get tension in the two cables in terms of W. Set the tension = 340 and solve for W, and select the smaller value of W that satisfies these conditions.
It may help to draw a free body diagram - remove the wall and replace it with force vectors acting longitudinally at the ends of the cables (in tension) and the strut (compression).
A Concurrent force system in space...
The difficulty in this topic lies is the coordinate distances as some people are easily confused, Three dimensional visualization is essential.
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The resultant of mutually perpendicular forces is given by the expression:
F= √[(Fx)² + (Fy)² + (Fz)²]
and its direction cosines by
cosϴx=(Fx/F)
cosϴy=(Fy/F)
cosϴz=(Fz/F)
The most convenient relation between the force and its components is expressed by
(F/d) = (Fx/x) = (Fy/y) = (Fz/z)
where x, y, z are the coordinate distance between points on the line of action and d= √(x²+y²+z²)
Solution:
Applying the above formulas
(AB/√42.25) = (ABx/2) = (ABy/6) = (ABz/1.5)
(AC/√49) = (ACx/2) = (ACy/6) = (ACz/3)
(AD/√42.25) = (ADx/0) = (ADy/6) = (ADz/2.5)
Consider Point A and apply method of joints, since you requires no tensile force shall exceed 340 ,set cables AB and AC = 340 lb
Summation of forces along Z axis:
[ΣFz = 0] ABz + ACz + ADz = P
(1.5*AB/√42.25) + (3*AC/√49) + (2.5*AD/√42.25) = P
(1.5*340/√42.25) + (3*340/√49) + (2.5*AD/√42.25) = P
224.1758242 + AD(5/13) = P............................equation (1)
Summation of forces along Y axis
[ΣFy = 0]
ABy + ACy - ADy = 0
(6*AB/√42.25) + (6*AC/√49) - (6*AD/√42.25) = 0
(6*340/√42.25) + (6*340/√49) - (6*AD/√42.250 = 0
605.2747253 - AD(12/13) = 0..........................equation (2).
Solving simultaneously;
P = 476.3736264 lb
AD = 655.7142857 lb (Strut is under Compression, It can be observed by Inspection)
Therefore it can be concluded that a weight of 476.3736264 lb will cause tension not more than 340 lb for cables AC and AB.
properly.. the unfastened physique diagram is effective however the Arrow Ay is backward (yet, it doesnt count, for the reason which you´ll have a adverse answer) And interior the case of the two toddlers the arrows must be pointing down, for the reason which you're saying that they are lifting the board. stable luck
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