Answer:
Therefore, at the highest or at maximum height, the net velocity is u= vocosθ. So, we can say that at the maximum height, the kinetic energy is minimum as the vertical velocity is zero.
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.
<span>The four forces are as follows: </span>
<span>Weight of the Box, down. (Let's call this W) </span>
<span>Tension of AB, toward B. (We'll call it Tb) </span>
<span>Tension of AC, toward C. (We'll call this Tc) </span>
<span>"Tension" from AD, from D. (And we'll call this D) </span>
<span>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>. </span>
<span>Breaking the force vectors into their components, we are left with the following equations: </span>
<span>(1) Fx = 0 = -4/13*Tb + 2/7*Tc </span>
<span>(2) Fy = 0 = -12/13*Tb + -6/7*Tc + 12/13*D </span>
<span>(3) Fz = 0 = 3/13*Tb + 3/7*Tc + 5/13*D - W </span>
<span>From (1), we can solve for Tb in terms of Tc, such that Tb = 13/14*Tc </span>
<span>From (2), we can substitute our solution from (1) into Tb and then solve D in terms of Tc, D = 13/7*Tc </span>
<span>Then from (3), we can substitute (1) and (2) for Tb and D and put W in terms of Tc, W = 19/14*Tc. </span>
<span>From (1), we can see that Tb will be less than Tc, so Tc shall be equal to 340 lbs. </span>
<span>W thus shall equal 460 lbs </span>
<span>D will equal 630 lbs</span>
p=F/A
or,P=d×V×G/A (m=d×V)
or,p= d× A×h×g/A (A and A are cut)
or,P=d×H×G
