A plane, which is flying horizontally at a constant speed Vo and at a height h above the sea, must drop a bundle of supplies to a castaway on a small raft. (a) Write down Newton's second law for the bundle as it falls from the plane, assuming you can neglect air resistance.
When the bundle falls, the only force acting on it at the time is gravity, since we are not taking into account air resistance. Therefore the force in the y direction = Fy= -mg. Since there is no horizontal movement, the x force Fx = 0. The net force on the bundle is F = -mgj or in vector form ( 0 , -mg ). From Newton's second law we have mx'' = ( 0 , -mg ) which simplifies to x'' = ( 0 , -g ).
Solve initial value problem for y(t) by Laplace Transformation. y' + 5.2y = 0, y(0)=1
1) Apply Laplace transform to differential equation with Linearity property so we get: L(y') + 5.2L(y) = L(1) The resulting transform of that equals (sY - y(0)) + 5.2Y = 1/s.* 2) Factor out the 'Y' term to get Y(s + 5.2) = 1/s, and then solve for Y to get: Y = 1/( s(s + 5.2) ) 3) Use partial fractions to break up the fraction into 1/( s(s + 5.2) ) = A/s + B/(s+2) which then simplifies to 1 = [A(s+2)]/s + (Bs)/(s+2). Simplifying again -> 1 = [(As + 2A) + Bs] / [ s(s+2) ] 4) Set the coefficients of the terms of the same order to then solve for A and B, so we obtain: A + B = 0 (for order of 1 terms) and 2A = 1 (for order of zero term) so then you get A = 1/2, and thus B = -1/2. 5) Plug the A and B terms back into the first partial fraction you made: Y = (1/2)/s + (-1/2)/(s+2) or Y = 1/(2s) - 1/2(s+2). 6) These terms are now in a manageable form to take the inverse Laplace transform of each of them to be in the form: L^-1(Y) = L^-1 [ 1/(2s)] - L^-1[ 1/2(s+2) ]. Taking the inverse laplace transform* we get: y(t) = (1/2)t - (1/2)e^-2t which is the final solution to the initial value problem. *Laplace transforms and inverse Laplace transforms are generally determined from a provided table of given transforms.
The cable will fail when subjected to a tension of 2kN. Determine the largest vertical load P the frame. http://mgh-images.s3.amazonaws.com/9780133918922/714361-7-10IP1.png
1) FREE BODY DIAGRAM: The first step in any statics or mechanics problem is to draw a free body diagram. Use the right-hand rule and define your axes. From this diagram, there will be the load force P pointing down at the left end of the rod, a tension force T, drawn from the top of the pulley to the right, and then finally the two x and y reaction forces (Ax and Ay) from the wall axle pointing towards point A. 2) REACTION FORCE EQUATIONS: The second step is to solve for the reaction forces by summing the forces in the x-direction to zero. From your free body diagram and the direction of your forces, you will see that T-Ax=0, so therefore T=Ax. Since we know the tension force is 12kN, Ax=12kN. Then you will repeat this step except with summing the y-forces to zero, so you will see that -P+Ay=0, so therefore P=Ay. 3) MOMENT EQUATIONS: Since you cannot solve for the load with the two force equations, you will need to write your third equation: Solving the moment around point A. The third equation is necessary because we have three unknown variables, and therefore we need three equations. We will choose to take the moment around point A. Why? This is because moment calculation is force times the perpendicular distance away from the point of reference. Therefore, we choose to take the moment around A because we can momentarily ignore the reaction forces (Ax and Ay), since they are zero distance from the reference point A. So, taking the moment around A* we get P(3*0.75)-12(0.6)=0. Solving for P we get P=0.533kN, so therefore Ay=0.533kN. *when determining the signs for the moment equations you imagine the point you take the moment from is a "wheel". The force that creates the moment will cause this "wheel" to turn, due to its distance from the point, causing the rotation. Using your right hand, curl your fingers in the direction that the force is turning the "wheel" and point your thumb. With traditionally defined axes, when your thumb points "out" of the page, or towards you, then the force creates a positive moment. If you curl your hand and point your thumb going "into" the page, then the force has created a negative moment, which explains the force signs above.