If members and can support a maximum tension of and, respectively, determine the largest weight of the crate that can be safely supported. 300 lb 250 lb AC AB A C B 4 ft 4 ft 3 ft 3 Solutions 44918 1/21/09 4:25 PM Page 128. Aqua data studio serial keygen. Cables OD and OB and the strut OC, required to support the 50-kg crate. The spring OA has an un-stretched length of 0.8 m and a stiffnesskOA =1.2 kN /m. The force in the strut acts along the axis of the strut. Solution: Strategy: First draw the free body diagram and write the equilibrium equation for point O. Multiple-piece shipments with a total weight of 200 lbs. Or more can qualify. Average package weight should be 15–25 lbs. Oblivion setup exe download pc. Per package, 108' in length, 165' in length plus girth (L+2W+2H). If the 34-lb crate is originally at rest on the ground when determine the crate’s velocity when Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. T = 0, tF = (e2t ) lb, A A 9196205R1p0479-0512 6/5/09 3:54 PM Page 499 22.
The 150 Lb Crate Is Supported By Cables Per
1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg.
2. w = mg
3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again.
The answers are:
T1 = 1862 N , T2 = 1620 N but I have no clue how to arrive at them