W (b) F s. W (c) F s

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1 Forces - Unit Problem A student whose mass is 57 kg rides an elevator while standing on a bathroom scale whose readings are in units of Newtons. What is the scale reading if the elevator moves from the 3rd to the 5th floor (a) at constant velocity, (b) at a velocity which increases at the rate of 4.1 m, (c) at a velocity which decreases at 2 s the rate of 3.2 m? s Use your experience in elevators, however limited, to guess how the scale reading compares to the weight in each of the cases. Use =, >, or < to indicate the relationship. (a) F s W (b) F s W (c) F s W. Your answers will be compared to the results of Newton s 2nd Law later. 6.2 Recognize again that a problem concerning forces requires Newton s 2nd Law. What is the object to be isolated in the problem? Student Scale Elevator 6.3 Considering the elevator will tell us nothing about the scale reading because that reading depends upon the interaction between the scale and the student. To help us to decide whether the student or scale should be isolated, draw two diagrams, one showing all of the forces exerted on the scale, the other all of the forces exerted on the student. Use words to describe each of your vectors. Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

2 6 4 Forces - Unit Problem 6.1 A student whose mass is 57 kg rides an elevator while standing on a bathroom scale whose readings are in units of Newtons. What is the scale reading if the elevator moves from the 3rd to the 5th floor (a) at constant velocity, (b) at a velocity which increases at the rate of 4.1 m, (c) at a velocity which decreases at 2 s the rate of 3.2 m? s Can you now explain why the student is to be isolated in this problem instead of the scale? Write your answer. Forces exerted by the student's feet Student's wei gh t Scale's weight Force exerted by scale Force exerted by the el evator floor (a) Forces on the scale (b) Forces on the student 6.5 None of the forces in diagram (a) are known, so the force equation for the scale contains all unknowns. At least we can determine the student s weight in diagram (b). But wait. When the force equation for diagram (b) gives you the force exerted by the scale on the student s feet, How is that force related to the scale reading, and why?

3 Forces - Unit Problem A student whose mass is 57 kg rides an elevator while standing on a bathroom scale whose readings are in units of Newtons. What is the scale reading if the elevator moves from the 3rd to the 5th floor (a) at constant velocity, (b) at a velocity which increases at the rate of 4.1 m, (c) at a velocity which decreases at 2 s the rate of 3.2 m? s The magnitude of the force exerted by the scale on the student s feet is equal to the scale reading. The reason is Newton s 3rd Law: The force exerted by the feet on the scale (causing a scale reading in diagram(a)) is equal in magnitude, but opposite to the force exerted by the scale on the feet (diagram (b)).

4 6 6 Forces - Unit Problem 6.1 A student whose mass is 57 kg rides an elevator while standing on a bathroom scale whose readings are in units of Newtons. What is the scale reading if the elevator moves from the 3rd to the 5th floor (a) at constant velocity, (b) at a velocity which increases at the rate of 4.1 m, (c) at a velocity which decreases at 2 s the rate of 3.2 m? s 2 End Page

5 Forces - Unit Problem A 0.7 kg mass (m 1 ) on a frictionless, horizontal surface is connected by an inextensible, massless string to a 0.4 kg mass (m 2 ). The string passes over a massless, frictionless pulley so that mass m 2 hangs below the surface, as shown. Find (a) the tension in the string, and (b) the acceleration of each mass. 6.7 Because we must find the force (tension) exerted on either block by the string, the 2nd Law will be used to solve the problem. From Chapter 5, we know that useful equations arise only by considering the forces exerted on each block. Draw vectors to represent all such forces below, describing each one in words. m 1 m 2 Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

6 6 8 Forces - Unit Problem 6.2 A 0.7 kg mass (m 1 ) on a frictionless, horizontal surface is connected by an inextensible, massless string to a 0.4 kg mass (m 2 ). The string passes over a massless, frictionless pulley so that mass m 2 hangs below the surface, as shown. Find (a) the tension in the string, and (b) the acceleration of each mass. 6.8 W eight of m 1 String tension Force exerted by table String tension W eight of m 2 Why does the vector on the string attached to mass m 1 point to the right? Why does the vector on the string attached to mass m 2 point upward? Write your answers.

7 Forces - Unit Problem A 0.7 kg mass (m 1 ) on a frictionless, horizontal surface is connected by an inextensible, massless string to a 0.4 kg mass (m 2 ). The string passes over a massless, frictionless pulley so that mass m 2 hangs below the surface, as shown. Find (a) the tension in the string, and (b) the acceleration of each mass. 6.9 At the table top, the vector on the string points to the right because the string pulls m 1 to the right. Below the table top, the vector on the string points upward because the string pulls m 2 upward, even as m 2 falls.

8 6 10 Forces - Unit Problem 6.2 A 0.7 kg mass (m 1 ) on a frictionless, horizontal surface is connected by an inextensible, massless string to a 0.4 kg mass (m 2 ). The string passes over a massless, frictionless pulley so that mass m 2 hangs below the surface, as shown. Find (a) the tension in the string, and (b) the acceleration of each mass. End Page

9 Forces - Unit Problem m m 1 2 m 3 Mass m 1 (38 kg) is pulled up an incline by a string attached to two falling masses m 2 (15 kg), and m 3 (28 kg) as shown. For a coefficient of friction of 0.13 between mass m 1 and the incline, find the tension in each string, and the acceleration of each mass We surmise once again that Newton s 2nd Law will be useful in finding a problem solution. From past experience, we know that it will prove useful to write a force equation for each block. So begin by drawing vectors in the diagram below to represent the forces exerted on each block. Use words to describe each vector. 40 Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

10 6 12 Forces - Unit Problem m m 1 2 m 3 Mass m 1 (38 kg) is pulled up an incline by a string attached to two falling masses m 2 (15 kg), and m 3 (28 kg) as shown. For a coefficient of friction of 0.13 between mass m 1 and the incline, find the tension in each string, and the acceleration of each mass string 1 tension frictional force m ' s weigh t 1 force exerted by surfa ce o at 90 stri ng 1 tensi on m ' s weigh t 2 stri ng 2 tensi on m ' s weigh t 3 Check that your vector representing m 1 s weight points straight downward, as shown. Explain the directions of the vectors on string 2. Why does the frictional force point in the direction shown?

11 Vectors - Unit Problem m m 1 2 m 3 Mass m 1 (38 kg) is pulled up an incline by a string attached to two falling masses m 2 (15 kg), and m 3 (28 kg) as shown. For a coefficient of friction of 0.13 between mass m 1 and the incline, find the tension in each string, and the acceleration of each mass The top vector on string 2 points downward because the string pulls mass m 2 downward. The bottom vector on string 2 points upward because the string pulls mass m 3 upward. The frictional force points down the plane because m 1 s motion up the plane is opposed by the rubbing between block and plane.

12 6 14 Forces - Unit Problem m m 1 2 m 3 Mass m 1 (38 kg) is pulled up an incline by a string attached to two falling masses m 2 (15 kg), and m 3 (28 kg) as shown. For a coefficient of friction of 0.13 between mass m 1 and the incline, find the tension in each string, and the acceleration of each mass. End Page

13 Forces - Unit Problem A man pushes a file cabinet, exerting a horizontal force so that the cabinet moves at constant velocity. The man s mass is 95 kg, and that of the filing cabinet is 56 kg. For a coefficient of kinetic friction between the filing cabinet and the floor of 0.4, (a) what is the total force (magnitude and direction) exerted by the floor on the man? (b) What force is exerted by the man on the file cabinet? 6.13 Because we are asked to find a force, we will use Newton s 2nd Law. The two objects have been separated so that all forces can be described clearly. Draw vectors to represent the forces exerted on each object. Describe each vector in words. Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

14 6 16 Forces - Unit Problem 6.4 A man pushes a file cabinet, exerting a horizontal force so that the cabinet moves at constant velocity. The man s mass is 95 kg, and that of the filing cabinet is 56 kg. For a coefficient of kinetic friction between the filing cabinet and the floor of 0.4, (a) what is the total force (magnitude and direction) exerted by the floor on the man? (b) What force is exerted by the man on the file cabinet? 6.14 (a) (b) Frictional force exerted by floor Man's weight Perpendicular forces exerted by floor Forces exerted by cabinet Forces exerted by hands Frictional force exerted by floor Cabinet' s weight Perpendicular force exerted by floor Why is the frictional force on the man directed to the right, while the frictional force on the cabinet is directed to the left? Why does the frictional force not have the same direction in both cases? Write your answers The man moves to the right only because of the frictional force exerted by the floor; if he were standing on ice this force would be much smaller and he would likely not move at all. The frictional force, however, opposes the rightward motion of the cabinet, so it is directed to the left in diagram (b).

15 Forces - Unit Problem A wall is being painted by using a paintbrush of mass 0.4 kg. The painter exerts a force on the handle of the brush at an angle 50 above the horizontal. If the paint is to be applied smoothly, at constant velocity, (a) what force is exerted on the handle? (b) What force is exerted by the wall on the paintbrush? The coefficient of kinetic friction between the paintbrush and wall is The forces will be calculated by using Newton s 2nd Law. Obviously the paintbrush is the object to isolate. Draw vectors to represent all of the forces exerted on the paintbrush as it moves up the wall. Describe your vectors in words. Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

16 6 18 Forces - Unit Problem 6.5 A wall is being painted by using a paintbrush of mass 0.4 kg. The painter exerts a force on the handle of the brush at an angle 50 above the horizontal. If the paint is to be applied smoothly, at constant velocity, (a) what force is exerted on the handle? (b) What force is exerted by the wall on the paintbrush? The coefficient of kinetic friction between the paintbrush and wall is force of h and 50 o brush's weight frictional force perpedicular force exerted by wall Explain why the force exerted by the wall on the paintbrush is in the direction shown If the wall were made of a weak plaster, the paintbrush would push a hole in the wall. Because there is no such hole, it must be that the wall pushes to the left on the brush.

17 Forces - Unit Problem You wish to perform the trick of removing the tablecloth without touching the cup and saucer which rest upon it. If you pull with a large enough force on the tablecloth, it can be made to clear the table while the cup and saucer move very little. What is this minimum force? The cup and saucer together have a mass of 0.23 kg, and the tablecloth s mass is 0.19 kg. The coefficient of static friction between the saucer and tablecloth is 0.2; the coefficient of kinetic friction between table and tablecloth is Finding the minimum force requires Newton s 2nd Law. What is the object to be isolated? We isolate the tablecloth since we are interested in the minimum force to be exerted on it. Draw vectors, with labels in words, to represent the horizontal forces exerted on the tablecloth if it moves to the left. Workbook for Introductory Mechanics Problem-Solving Copyright 1996, 1997 by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

18 6 20 Forces - Unit Problem 6.6 You wish to perform the trick of removing the tablecloth without touching the cup and saucer which rest upon it. If you pull with a large enough force on the tablecloth, it can be made to clear the table while the cup and saucer move very little. What is this minimum force? The cup and saucer together have a mass of 0.23 kg, and the tablecloth s mass is 0.19 kg. The coefficient of static friction between the saucer and tablecloth is 0.2; the coefficient of kinetic friction between table and tablecloth is Force exerted by hand Force exerted by entire table top Force exerted by cup and saucer Notice that the table top, and the cup and saucer oppose the motion of the tablecloth, resulting in the vector directions shown.