500 N 2 m 2 m 2 m 3 500 N 200 N • m Find the equivalent system of forces about point O FR = ΣΕ - (Μ.) =ΣΜ +ΣΜ = 0
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We are given the forces acting and the moment about point O. We first add the forces to get the net force acting. We then add the torques or moments about point O to get the equivalent moment. We take counterclockwise as +ve direction and clockwise and negative direction.
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- A 2.220 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is us = 0.605 and the coefficient of kinetic friction is µ 0.255. At time t = 0, a force F 8.10 N is Us,k applied horizontally to the block. State the force of friction applied to the block by the table at times t = 0 and t > 0. t = 0 N t > 0 N Consider the same situation, but this time the external force F is 16.3 N. Again, state the force of friction acting on the block at times t = O and t > 0. t = 0 N t > 0 N * TOOLS x10A scale is constructed using the 10* kg massthe 2-kg pan Pand the pulley and cord arrangementCord BCA is 2 m long. Ifs s = 0.75m determine the mass D in the pan. Neglect the size of the pulleyWhen you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.30 kgkg . If the biceps is connected to the forearm a distance dbicepsdbicepsd_biceps = 3.50 cmcm from the elbow, how much force FbicepsFbicepsF_biceps must the biceps exert to hold a 600 gg ball at the end of the forearm at distance dballdballd_ball = 33.0 cmcm from the elbow, with the forearm parallel to the floor? How much force FelbowFelbowF_elbow must the elbow exert? Although it is uncommon, orthopedic surgeons have seen patients who have torn their biceps tendon when forces of greater than 390 NN have been exerted on the tendon with the arm bent at the elbow so that the forearm is parallel to the ground. What is the minimum mass of a ball MballMballM_ball such a patient might have held to cause the biceps tendon to tear?
- 3-43. The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium. Prob. 3-43 1.5 m 15 m 2 m O F AD = 763 N, F_AC = 392 N, F_AB = 523 N O FAD = 523 N, F_AC = 763 N, F_AB = 392 N O FAD = 392 N, F_AC = 523 N, F_AB = 763 NB h = 150 cm x = 50 cm L = 95 cm Mass = 0.8 kg at point G (so MMI = mL^2 about point A) At what acceleration (in m/s^2) will the object lift off the wall?The rectangular framework is subjected to the following non-concurrent system of forces: a 30lb- force below the origin, a 20lb-force 3ft above the origin and making an angle of 30°NE, a 100lb-force with a horizontal distance of 3ft and a vertical distance of 4ftfrom the origin and making an angle of 45° NW, a 60lb force with the same distance like the previous force but along the positive x-axis, and a 100lbs force 4ft to the right of the origin and making an angle of 60° SW. Determine the magnitude and direction of the resultant, as well as its moment arm relative to the origin.
- A 15 N horizontal force F→ pushes a block weighing 3.7 N against a vertical wall (see the figure). The coefficient of static friction between the wall and the block is 0.61, and the coefficient of kinetic friction is 0.42. Assume that the block is not moving initially. (a) Will the block move? ("yes" or "no") (b) In unit-vector notation Fxî+Fyĵ, what is the force on the block from the wall?A tutor told me that the answer to 1b was 10N but I don't understand why it would be 10N for the tension. If that is correct could someone break it down for me?A 2.220 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is µs = 0.605 and the coefficient of kinetic friction is µ = 0.255. At time t = 0, a force F = 8.10 N is m Us,k applied horizontally to the block. State the force of friction applied to the block by the table at times t = 0 and t > 0. t = 0 N t > 0 N Consider the same situation, but this time the external force F is 16.3 N. Again, state the force of friction acting on the block at times t = 0 and t > t = 0 N t > 0 N * TOOLS x10
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