Welcome back to PHY 211

1. Welcome back to PHY 211 Today’s Agenda Apparent Weight Forces in Circular Motion

2. A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is moving, the reading is frequently changing, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the elevator 1. must be going up 2. must be going down 3. could be going up or going down 4. I’m not sure.

3. Free body diagram for person There are two forces acting on the person: First, there is the force of gravity produced by the earth, i.e. the weight of the person. Second, there is the contact force produced by the scale acting on the person.

4. How do we know that there are no other forces on the person? Answer: The person is in contact only with the scale.

5. Motion of elevator (if a = ) Moving upward and slowing down, OR Moving downward and speeding up. Motion of elevator (if a = ) Moving upward and speeding up, OR Moving downward and slowing down.

6. Applying Newton’s Second Law • Fnet = ma • Fnet is the sum of all the forces on the person. Also m is the mass of the person and a is the acceleration. • Fnet is N – W, where N is the normal force produced by the scale and W is the weight of the person. • Combining, gives N – W = ma

7. Finding the Apparent Weight • The apparent weight Wapparent is simply the normal force N. We simply solve for N to get N = W + ma Hence the apparent weight exceeds W, if a is positive. This occurs, for e.g. when the elevator is rising with increasing speed

8. Continued discussion of Apparent Weight • Suppose the elevator is descending but with increasing speed • Then acceleration a is negative • We again apply our general result N = W + ma We see that the normal force now is less than the actual weight W.

9. Special Case – Free Fall • IF the elevator is in free fall, its acceleration is –g. Then what is the apparent weight? Again using N = W + ma, we see that N = W – mg = 0!!! i.e. The scale will read zero.

10. Conclusions • Scale reads magnitude of normal force N • Reading on scale does not depend on velocity • Depends on acceleration only • a>0 normal force bigger • a<0 normal force smaller

11. A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is accelerating, a different reading is observed, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e., 200 lbs) the acceleration of the elevator is approximately 1. 1 m/s2 2. 2.5 m/s2 3. 5 m/s2 4. 12.5 m/s2

12. Problem to Test your understanding A woman is in a descending elevator which is slowing down. It is losing 2.0m/s every second. The weight of the woman is 490N. Find her apparent weight on the scale on which she is standing.

13. Forces in circular motion

14. Review on Circular Motion An object of mass m is moving with uniform speed v in a circle The radius of the circle is r What is the direction and magnitude of the acceleration vector? Recall that the acceleration vector is directed radially inward The magnitude is v2/r

15. Net Force on the mass By Newton’s second law, there must be a net force on the mass which is directed radially inward. This is due to the fact that the velocity vector is changing direction. Applying Newton’s second law, the net force Fnet = ma = mv2/r

16. Example A mass m moves on a frictionless track in a vertical plane with constant speed v The radius of the track is r At the bottom of its path, find the normal force N

17. Solution The two forces acting on mass m are the normal force N acting upward and the weight force W acting downward Hence Fnet = N – W Equating to ma, we obtain N – W = ma = m(v2/r) It follows that the normal force N is N = mg + m(v2/r) Note that the normal force exceeds the weight