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Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.5N. From FBD(a), + c ©Fy = 0; N + P sin 20° - 80(9.81) = 0 (1) + ©F = 0; : x. P cos 20° - 0.5N = 0 (2) Determine the force in each member of the space truss and state ... Determine the force required to hold the 150-kg crate in equilibrium. 6-2-24. p.314, 6-83.

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+2 Partially Correct Equilibrium Equations +1 Equilibrium Equations with Major Errors lution +1 Incorrect Use of Max Friction Form 9.9/14 (70.7%) {ml) kg {m2) kg A crane is lifting two crates. Part Description Determine the maximum force T that can be applied so the cable does not break. (include units with answer) ml=51 1112=75 Answer: 1500 N Let crate 1 be the \(\text{50}\) \(\text{kg}\) crate and crate 2 be the \(\text{30}\) \(\text{kg}\) crate. We will choose up the slope as positive. We need to find all the forces that act parallel to the slope on each box in turn and use this to solve simultaneously for the frictional force. We draw free body diagrams for each crate:

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the weight of the 30-kg cylinder. 150 30 kg 300 3/10 Determine the force magnitude P required to lift one end of the 250-kg crate with the lever dolly as shown. State any assumptions. 250 kg 1500 mm 275 mm

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A 150-kg crate is released from an airplane traveling due east at an altitude of 7400 m with a ground speed of 120 m/s. The wind applies a constant force on the crate of 250 N directed horizontally in the opposite direction to the plane's flight path.

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Determine the displacement y of block A for equilibrium if the masses of blocks A and B are 22 kg and 34 kg, respectively. SOLUTION From a free-body diagram of the hanging weight, the equations of equilibrium n 6Fy 0: T 34 9.81 0 are solved to get T 333.54 N Since the pulleys are free to rotate, the tension is the same in every part of the rope. In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate.