A box with mass m=1.25 kg rests on the top of a bar. The coefficient of static friction between the box and the bar is μs=0.73 and the coefficient of kinetic friction is μk=0.39. a) Write an expression for Fm the minimum force required to produce movement of the box on the top of the bar. b) Solve numerically for the magnitude of the force Fm in Newtons. c) Write an expression for a, the box's acceleration, after it begins moving. (Assume the minimum force, Fm, continues to be applied.) d) Solve numerically for the acceleration, a in m/s².

A box with mass m=1.25 kg rests on the top of a bar. The coefficient of static friction between the box and the bar is μs=0.73 and the coefficient of kinetic friction is μk=0.39. a) Write an expression for Fm the minimum force required to produce movement of the box on the top of the bar. b) Solve numerically for the magnitude of the force Fm in Newtons. c) Write an expression for a, the box's acceleration, after it begins moving. (Assume the minimum force, Fm, continues to be applied.) d) Solve numerically for the acceleration, a in m/s².

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4. You are pushing horizontally with a 300 N force on a box that is on an inclined plane as shown. The mass of the box is 30 kg and the incline is at a 25 degree angle off the Fapplied horizontal. The coefficient of kinetic friction between the box and the incline is 0.3, and the coefficient of static friction i 0.4. Assuming the box was at rest at t=0 when the force is initially applied, is the box moving up the ramp, down the ramp, or is it at rest after the force has been applied? 25⁰
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GOAL Apply the concept of static friction to an object resting on an incline. PROBLEM Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is ng sin e 0.350, what maximum angle can the ramp make with the horizontal mg cos e e before the block starts to slip down? STRATEGY This is an application of Newton's second law involving an object in equilibrium. Choose tilted coordinates, as in the figure. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, f. = 4n. SOLUTION Write Newton's laws for a static system (1) EF = mg sin e - u̟n = 0 in component form. The gravity force (2) EF = n - mg cos e = 0 has two components. Rearrange Equation (2) to get an n = mg cos 8 expression for the normal force n. Substitute the expression for n into EF, = mg sin 8 - H mgcos 0 = 0 → tan 8 =u. Equation (1) and solve for tan 0. Apply the inverse tangent function to get the answer.…
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The tolerance is -+1 in the 2nd significant digit