COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 5, Problem 39QAP
To determine
The minimum horizontal force that will cause an object resting on a level floor to start sliding up.
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8-63. Determine the smallest force P that will cause
impending motion. The crate and wheel have a mass of
50 kg and 25 kg, respectively. The coefficient of static
friction between the crate and the ground is , = 0.2, and
between the wheel and the ground, = 0.5.
*8-64. Determine the smallest force P that will cause
impending motion. The crate and wheel have a mass of
50 kg and 25 kg. respectively. The coefficient of static
friction between the crate and the ground is , = 0.5, and
between the wheel and the ground μ = 0.3.
O
O
O
127
10
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C
A
300 mm
Estimate the height of a typical playground slide and theangle its surface makes with the horizontal. A child wearing a different sort of clothing than the first child climbs the ladder to the top of the slide, sits on the slide, lets go of the handrail, and remains at rest. What is the minimum possible value for the coefficient of static friction between this child and the surface of the slide?
Question 7
A 81.17 kg crate is on a frictionless surface inclined at an angle of 0 = 28.64°. A constant external force, P, is applied horizontally to the crate
such that it moves a distance of 3.00 m up the incline at constant speed. What is the magnitude of the force P? (740)
%3D
P.
3.0 m
A) O1303.27 N
B) O381.27 N
C) 0434.42 N
D) O14.09 N
E) 698.14 N
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Chapter 5 Solutions
COLLEGE PHYSICS
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- A uniform pole is propped between the floor and the ceiling of a room. The height of the room is 7.80 ft, and the coefficient of static friction between the pole and the ceiling is 0.576. The coefficient of static friction between the pole and the floor is greater than that between the pole and the ceiling. What is the length of the longest pole that can be propped between the floor and the ceiling?arrow_forwardTo determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 36.0 and then moves down the incline with constant speed when the angle is reduced to 30.0. From these data, determine the coefficients of static and kinetic friction for this experiment.arrow_forwardA box of mass m rests on a rough, horizontal surface with a coefficient of static friction s. If a force Fp is applied to the box at an angle as shown, what is the minimum value of for which the box will not move regardless of the magnitude of Fp?arrow_forward
- An automobile driver traveling down an 8% grade slams on his brakes and skids 30 m before hitting a parked car. A lawyer hires an expert who measures the coefficient of kinetic friction between the tires and road to be k = 0.45. Is the lawyer correct to accuse the driver of exceeding the 25-MPH speed limit? Explain.arrow_forwardIf you hold a horizontal metal bar several centimeters above the ground and move it through grass, each leaf of grass bends out of the way. If you increase the speed of the bar, each leaf of grass will bend more quickly. How then does a rotary power lawn mower manage to cut grass? How can it exert enough force on a leaf of grass to shear it off?arrow_forward(a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00105 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.17. The strand sags at an angle of 12° below the horizontal. Compare this with the tension in the vertical strand (find their ratio).arrow_forward
- Lisa measured the coefficient of static friction between two pairs of running shoes and the track in Example 6.1 (page 159). If she wants to have an advantage in a race, which shoes should she wear, the ones with a high coefficient or the ones with the low coefficient of static friction? Explain.arrow_forwardA block of ice (m = 15.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal force of 95.0 N for 1.54 s. a. Determine the magnitude of each force acting on the block of ice while you are pulling. b. With what speed is the ice moving after you are finished pulling? Repeat Problem 71, but this time you pull on the block at an angle of 20.0.arrow_forwardConsider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?arrow_forward
- This problem returns to the tightrope walker studied in Example 4.6, who created a tension of 3.94103 N in a wire making an angle 5.0° below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter.arrow_forwardA truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the acceleration of the truck if its trailer leaks sand at a constant rate through a hole in its bottom? (a) It decreases at a steady rate. (b) It increases at a steady rate. (c) It increases and then decreases. (d) It decreases and then increases. (e) It remains constant.arrow_forward4) A 30-kilogram child is sitting 2.0 meters from the centre of a merry-go-round. The coefficients of static and kinetic friction between the child and the surface of the merry-go-round are 0.8 and 0.6, respectively. Determine the maximum speed of the merry-go-round before the child begins to slip.arrow_forward
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Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY