PH 201

Dr. Cecilia VogelLecture 8

REVIEW

ForcesVarious types of forcesFree Body Diagrams

Gravitynormaltensionspring

OUTLINECentripetal forcessprings2-D problems

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!

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

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

k is called the “spring constant,” depends on the spring’s stiffness.

Fspring = -kx

F

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?

spring,1 1

1.0N (3.0mm)

F k x

k

k = 0.333 N/mm

Given: F1 = 1.0 N, so Fspring,1 = -1.0N,x1 = 3.0mm

F2 = -9.8 N, so Fspring,2 = 9.8 N

spring,2 2F k x Use: but need k.

First, find k:

Want x2 =?

So:29.8N (0.33N/mm) x x2 = -29.4

mm

Fooling the Scale – Case 3A 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

Fooling the Scale – Case 3What is the angle of the slope?

F ma Now can we apply 2nd law?

weight

N

Useful x- and y-axes are along slope and perpendicular.

xy

y yF ma x xF ma

Let’s consider the y-component

What forces are in the y-direction?Normalone component of gravity

Fooling the Scale – Case 3

cosN mg ma

2931N (110kg)(9.8m/s )cos 0

30o

weight

N x

y

y yF ma What forces are in the y-direction?Normalone component of gravity

this component is given by

What is the accel in the y-direction?zero

mgcos