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Description: A 3 kg box sits on a 35 degree incline. Calculate the minimum value for the coefficient of friction if the box is not moving. Indicate all forces on a diagram.

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Physics

A 3 kg box sits on a 35 degree incline. Calculate the minimum value for the coefficient of friction if the box is not moving. Indicate all forces on a diagram.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter6: Applications Of Newton’s Laws Of Motion

Section: Chapter Questions

Problem 11PQ: In Problem 10, the mass of the sign is 25.4 kg, and the mass of the potted plant is 66.7 kg. a....

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In Problem 10, the mass of the sign is 25.4 kg, and the mass of the potted plant is 66.7 kg. a. Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the potted plant. b. What is the maximum value of the static friction force if the coefficient of static friction between the pot and the roof is 0.572?
It is observed that when the bed of the dump truck is raised to an angle of u = 20 degree the vending machines will begin to slide off the bed as shown in the figure below. Determine the static coefficient of friction between a vending machine and the surface of the truckbed.
Of the following statements, indicate on the line whether they are true (T) or false (F):___ The friction between two surfaces can be changed by changing the masses of the objects.___ The coefficient of static friction can be less than the coefficient of kinematic friction.___ When the force of friction increases, the coefficient of friction decreases.___ The only way to change the friction between two objects is by changing their masses.___ The only way to change the friction between two objects is to polish their surfaces.___ The only way to change the friction between two objects is by changing their materials.___ The value of the friction force of a body is always constant when it is staticand does not vary with the applied force.___ When a ball strikes outside the center of gravity of a baseball bat, the ball will travelfurther than if it hit the center of gravity.___ The center of gravity of a body may be outside of it.___ The center of gravity of an object depends only on the…
A crate of mass 15.0 kg rests on a level surface. If the coefficient of kinetic friction between the crate and surface is 0.440, calculate the following. (a) Calculate the normal force. (Give the magnitude.) N(b) Calculate the magnitude of the kinetic friction force when a horizontal applied force of 67.0 N moves the crate. N(c) Calculate the normal force when the 67.0-N applied force is exerted at an angle of 37.5° above the horizontal. (Give the magnitude.) N(d) Calculate the magnitude of the kinetic friction force when the 67.0-N applied force is exerted at an angle of 37.5° above the horizontal. N
Which of the following statements is true about friction? k and us are independent on the mass of the object. The coefficients of friction are independent on the nature of the surfaces. Hk is greater than us Static friction is relatively constant while kinetic friction is not. The force of static friction depends on the area of contact between the two surfaces.
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A 5.00-kg box is initially at rest on a level surface. The coefficient of static friction between the box and the surface is 0.400. Calculate the maximum force you can apply before the box starts to move, if the force is directed at an angle of 30 degrees from the horizontal, with the vertical component downward. Calculate the maximum force you can apply before the box starts to move, if the force is directed at an angle of 30 degrees from the horizontal, with the vertical component upward.
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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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A 50-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 30.0 degrees below the horizontal is applied to the box. The coefficients of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.30. What is the force of friction on the box? Draw force diagram. Please provide step-by-step solution