Welcome back to Physics 211

Welcome back to Physics 211 Today’s agenda: Friction – block on plane Internal forces Tension Circular motion

Announcements • MP Homework 3 due Friday 12 pm

Block on plane revisited F N W q

Initially at rest • What is largest angle before slips ? Resolve perpendicular to plane  N=Wcosq resolve parallel  F=Wsinq since F <= mN we have Wsinq <= mWcosq ie tanq <= m

Angle > tan-1m • Resolve along plane Wsinq-mKWcosq=a Or a=g(sinq-mKcosq)

Friction demo • Static friction: depends on surface and normal force for pulled block • Kinetic friction generally less than maximal static

Summary of friction • 2 laws of friction: static and kinetic • Static: friction tends to oppose motion and is governed by inequality Fs <=msN • Kinetic friction is given by equality FK=mKN

Internal Forces • So far replaced macroscopic bodies by points – why is this ok ? • Specifically, such body composed of (very many) parts – neglected all internal forces of these parts on each other • Also neglected rotational motion -- later

Simple example A B Const v B: A: NAG NBG NAB PAH NBA FAG FBG WBE WAE

Newton’s 2nd Law Newton’s 2nd law (NSL) for A (a=0): NAG+WAE=0; PAH+NAB+FAG=0 NSL for B: NBG+WBE=0; NBA+FBG=0 add  NAG+NBG+WAE+WBE=0; PAH+(NAB+NBA)+FAG+FBG=0

Newton’s 3rd law NAG+NBG+WAE+WBE=0 N (A+B),G+W(A+B),G=0 and PAH+(NAB+NBA)+FAG+FBG=0  PAH+F(A+B),G=0

Composite system NTL for A and B imply that can consider combined system C=A+B in which NAB etc do not appear – internal forces NCG=NAG+NBG PCH FC=FAG+FBG WCE=WAE+WBE

Internal forces summary • Can apply N’s laws to composite body • Can ignore internal forces of one part of body on another since cancel (NTL) • Justifies treating macroscopic bodies as point-like …

Tension

Two blocks are connected by a heavy rope. A hand pulls block A in such a way that the blocks move upward at increasing speed. The (downward) tension force on the upper block by the rope is 1. less than 2. equal to 3. greater than the (upward) tension force on the lower block by the rope. 4. Answer depends on which block is heavier.

Hand pulls block A so blocks move up at increasing speed.

Notice: for mR=0, the tension forces exerted at either end are the same. The term “tension in the string” is therefore often used as a short-hand for the tension forces exerted on or bythe string at either end.

Blocks A and C are initially held in place as shown. After the blocks are released, block A will accelerate up and block C will accelerate down. The magnitudes of their accelerations are the same. Will the tension in the string be 1. equal to 1.0 N (i.e., the weight of A), 2. between 1.0 N and 1.5 N, 3. equal to 1.5 N (i.e., the weight of C), or 4. equal to 2.5 N (i.e.,the sum of their weights)?

Free-body diagram for block A. Free-body diagram for block C.

Pulleys etc 2 pulleys 2T=W F=T=W/2 T N pulleys F=W/N! F W

Forces in circular motion .

Two identical balls are connected by a string and whirled around in circles of radius r and 2r at constant speed. The acceleration of ball B is 1. four times as great 2. twice as great 3. equal to 4. one half as great as the acceleration of ball A.

The two balls are whirled around in a circle as before. Assume that the balls are moving very fast and that the two strings are massless. The tension in string P is 1. less than 2. equal to 3. greater than the tension in string R.

Motion of car on banked circular track car N a= q R W Speed v Horizontal forces: Vertical:

Motion on loop-the-loop what is normal force on car at top and bottom of loop ? Neglect friction Assume moves with speed vB at bottom and vT at top car

Free body diagrams At bottom At top Newton:

Apparent weight ? What is criteria to just make it over loop ?

Another example m2 falls coefficient of kinetic friction on plane is m m1 m2 q

Free body diagrams Block 2 Block 1

Newtons 2nd law:

Criteria for Motion ? What is largest value of m which supports motion ? (assume q=300, m1=1.0 kg, m2=2.0 kg)

Welcome back to Physics 211