Statics of Rigid Bodies pros. I badly need your help, please?

1. If a 200 kN load is hanging from a ceiling by means of rope, which makes a 70 degrees angle with the ceiling, determine the horizontal force H responsible for the inclination, and the tension T in the rope.2. The unstretched length of an elastic cord is made to run vertically from the ceiling down to the floor. If the mid-portion of the elastic cord is pulled with a 15 kN horizontal force, thus stretching the cord and making the cord incline with a slope of 4/5, determine the tension in the cord.3. A concurrent force system consist of three forces with magnitude of 120,50 and 130 kN. What are the angles among them assuming that the force system is in equilibrium?4. A 20 kg ball snuggly fits inside a cube, whose bottom is tipped at 45 degrees from the horizontal. Determine the reaction of the bottom of the cube due to the weight of the ball. Assume forces are smooth.Having a hard time and I need an idea. I'm not sure of what I have.. Share your knowledge please.. Thank you!

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The vertical forces are in balance, so200 kN = T cos(70) andT = 200 kN/cos(70) = 585 NH = T sin(70) = 549.5 NCord problem:15 kN = T (3/5) => T = 25 kNThree concurrent forces:Without loss of generality, position the 130 kN force F1 along the positive x-axis.The 120-kN force F2 must then point somewhere in the 2nd or 3rd quadrant;it doesn't matter which we assume, the difference will just be a mirror image.So assume F2 points into the 2nd quadrant.The 50-kN force F3 must now point into the 3rd quadrant.Let A2 be the angle between F1 and F2; let A3 be between F1 and F3.50 cos(A3) + 120 cos(A2) + 130 = 0..........."equation (1)"50 sin(A3) = 120 sin(A2)A3 = arcsin[2.4 sin(A2))Therefore, cos(A3) = sqrt[1 - 5.76 sin^2(A2)]Plug this in to "equation (1)" and you have an equation whoseonly unknown is A2. Also note that cos(A2) = sqrt[1 - sin^2(A2)].Then the only unknown in "equation (1)" will be sin(A2),and the equation can be solved by algebra alone (and a trig table).Ball: I guess they mean the cube now rests on one edge.The weight of the ball is then exactly "split" between the two cube facesupon which it rests. Answer: (1/2)mg = 98 N.