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PH 201 Dr. Cecilia Vogel Lecture 8. REVIEW  Forces  Various types of forces  Free Body Diagrams  Gravity  normal  tension  spring OUTLINE  Centripetal.

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Presentation on theme: "PH 201 Dr. Cecilia Vogel Lecture 8. REVIEW  Forces  Various types of forces  Free Body Diagrams  Gravity  normal  tension  spring OUTLINE  Centripetal."— Presentation transcript:

1 PH 201 Dr. Cecilia Vogel Lecture 8

2 REVIEW  Forces  Various types of forces  Free Body Diagrams  Gravity  normal  tension  spring OUTLINE  Centripetal forces  springs  2-D problems

3 Centripetal force  Centripetal force  any force that causes centripetal acceleration  might be tension, gravity, friction…  force toward center  Centripetal force in lab  the string is pulling the cart toward the center of rotation  why doesn’t cart accelerate toward center?  IT DOES!!  any object moving in circle is accelerating toward center!

4 Centripetal force  Centripetal force in lab  when cart is pinned to end of track  centripetal force provided by tension in string and normal force of pin  when the cart is about to move toward center, centripetal force = tension in string only  This is the instant you analyze  after the cart starts moving toward the center  it’s no longer uniform circular motion  no longer just centripetal accel

5 Springs  If a spring is stretched or compressed, there will be a restoring force.  The force that the spring exerts is opposite the displacement,  and proportional to the size of the displacement. PAL xx  k is called the “spring constant,” depends on the spring’s stiffness. F spring = - k  x F

6 Spring Constant Example  If a 1.0 N force stretches a particular spring by 3.0 mm, how much will that spring stretch with a 1.0 kg-weight hanging from it vertically? k = 0.333 N/mm Given: F 1 = 1.0 N, so F spring,1 = -1.0N,  x 1 = 3.0mm F 2 = -9.8 N, so F spring,2 = 9.8 N Use:but need k. First, find k: Want  x 2 =? So:  x 2 = -29.4 mm

7 Fooling the Scale – Case 3 A 110-kg man stands on a scale. He and the scale are on a cart that is rolling down an incline. The scale reads only 95 kg again. What is the angle of the incline from horizontal? Use a free-body diagram to help solve this problem. What is our object? the man What are all the forces acting on the man? gravity & normal force from scale always perpendicular to surface weight N

8 Fooling the Scale – Case 3 What is the angle of the slope? Now can we apply 2 nd law? weight N Useful x- and y-axes are along slope and perpendicular. x y Let’s consider the y-component What forces are in the y-direction?  Normal  one component of gravity

9 Fooling the Scale – Case 3 weight N x y What forces are in the y-direction?  Normal  one component of gravity  this component is given by What is the accel in the y-direction? zero mg cos 


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