College Physics
1st Edition
ISBN: 9781938168000
Author: Paul Peter Urone, Roger Hinrichs
Publisher: OpenStax College
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Chapter 5, Problem 12PE
If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is
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If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is θ = tan–1 μs. You may use the result of the previous problem. Assume that a=0 and that static friction has reached its maximum value.
A block is resting on a ramp as shown in the figure below. You can change the inclination angle θ by raising one end of the ramp. The block has a mass mb = 8.5 kg. At the interface between the ramp and the block, the coefficient of static friction is μs = 0.450, and the coefficient of kinetic friction is μk = 0.270.If you raise the end of the ramp very slowly, at what angle will it slip?
A student of physics is carrying a heavy wooden crate. Being educated about friction, this student temporarily reduces the burden by pressing the crate against a wooden wall with a force, Fapp, applied at an angle θ above the horizontal, as shown in the figure. Assume that the crate slides with a small but constant speed up the wall to a point where the student can grip it more comfortably. The coefficients of static and kinetic friction are μs and μk, respectively. Coordinate axes have been indicated. Please use the interface below to create your free body diagram for the crate. If appropriate, use Fg to represent a weight, use Fn to indicate a normal force, use Fs to indicate static friction, and use Fk to represent kinetic friction. Enter an expression for the x component of the net force on the crate in terms of the various forces acting upon it. Part (c) Enter an expression for the y component of the net force on the crate in terms of the various forces acting upon it. Part (d)…
Chapter 5 Solutions
College Physics
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- The weight of the box is W = 90 N and the coefficient of static friction between the box and the floor is μ=0.65. Neglect the weights of the bars. What is the largest value of the force F that will not cause the box to slip?arrow_forwardA man weighing 400N climbs up a ladder 5 meters long. The ladder initially rests against a frictionless vertical wall and is at 58° above a rough floor with static friction coefficient of 0.25. determine the location of the man that will cause the ladder to slide down. What is the magnitude and direction of the net force acting on the ladder at the floor? FIND: location of man to cause sliding measured along the ladder (in meters) magnitude of the Net Force on the ladder at the wall (in Newton) angle and direction of the Net force on the ladder at the wall (degrees and N-E-W-S)) Magnitude of the Net force on the ladder at the floor(in Newton) angle and direction of the Net force on the ladder at the floor(degrees and N-E-W-S))arrow_forwardA box is placed at one end of a 2.00-m plank. That end of the plank is slowly raised until the box begins to slide. How much time does it take the box to slide to the other (bottom) end of the plank? The coefficients of static and kinetic friction between the box and the plank are, respectively, μs=0.400 and μk=0.200.arrow_forward
- A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that you can obtain the coefficient of static friction μs by measuring the critical angle θ where slipping occurs. Solution Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: ΣFx=mg-f= Evaluating, we get: f = mg But the frictional force is expressed as the product of coefficient and the normal force so. μs = mg For an inclined plane. The normal force is n=mg Then, μs=mg/cos(θ) so: μs=tan(θ)arrow_forwardThe weightless strut in the figure below is not attached to the wall; it is prevented from falling only by friction. (Let w = 375 N, L = 5.05 m and h = 3.10 m. (a) Find the magnitude of the force of friction between the wall and the strut.(b) Find the normal force exerted by the wall on the strut.(c) Find the minimum coefficient of static friction. please solve thank youarrow_forwardA ladder of length L and mass m leans against the side of a house, making an angle of θ� with the vertical. Assume that the ladder is free to slide at the point where it touches the side of the house (there is no significant friction). Find an expression for the magnitude of the normal force that the side of the house exerts on that end of the ladder in terms of m, g, L, and θ� . Evaluate the result for a ladder of mass 34 kg, a length of 4 m, and an angle of 25o from the vertical.arrow_forward
- A block of mass M = 10 kg is sitting on a surface inclined at angle θ = 45°. Given that the coefficient of static friction is μs = 0.5 between block and surface, what is the minimum force F necessary to prevent slipping? What is the maximum force F that can be exerted without causing the block to slip?arrow_forwardYou can hold a book against the wall, preventing it from sliding down, if you press your hand against it. Suppose the coefficient of static friction between the book and the wall is 0.450, but there is no friction between your hand and the book. The mass of your book is 3.75 kg. (a) What minimum force do you need to apply to keep the book from slipping down the wall? Give your answer in units of Newtons. Assume you are applying the force horizontally. (b) Now suppose you are pushing the book at an angle of 37.5 ̊ above the horizontal. Now what is the minimum force you need to apply to keep the book from slipping down the wall? Give your answer in units of Newtons.arrow_forwardA man weighing 400N climbs up a ladder 5 meters long. The ladder initially rests against a frictionless vertical wall and is at 58° above a rough floor with static friction coefficient of 0.25. Determine the location of the man that will cause the ladder to slide down. What is the magnitude and direction of the net force acting on the ladder at the floor?arrow_forward
- Consider a 25-kg block on a frictionless inclined plane held in place by a steel cable as shown in the figure. The cable is parallel to the surface of the incline. What is the magnitude of the tension on the steel cable? 25 kg 30° 212 N O 245 N O none of the choices O 123 N O 458 Narrow_forwardProblem: A block is resting on a ramp as shown in the figure below. You can change the inclination angle θ by raising one end of the ramp. The block has a mass mb=6.5 kg. At the interface between the ramp and the block, the coefficient of static friction is μs=0.4, and the coefficient of kinetic friction is μk=0.28. I managed to calculate that: slip angle is θ= 21.8 degrees acceleration is 1.09 m/s^2. I am trying to find: Let the angle θ be the angle you calculated , θslip (21.8 degrees) . To keep the block from slipping, you apply a horizontal (parallel to the surface of Earth) force directed to the right on the block. Your horizontal force F→Yb has a magnitude of 45 N. Calculate the magnitude and direction of the force due to static friction for this new situation, assuming that the block does not slip. Enter a magnitude for this part, but consider what the sign of your answer tells you about the direction of the force.arrow_forwardCalculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that force as the only unknown. The forces are positive if in tension, negative if in compression. 3.0 m BC= Answers: BE= A 3.0 m EF= G i i i B 3.0 m 3.0 m F 1.9 m E 3.0 m 20.0 KN KN KN kN Darrow_forward
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