PHY PHYSICS 231 Lecture 8: Forces, forces & examples Remco Zegers Walk-in hour: Monday 11:30-13:30 Helproom

PHY Newton’s Laws  First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity.  Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma  If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F 12 =-F 21

PHY Forces seen in the previous lecture Gravity: Force between massive objects Normal force: Elasticity force from supporting surface  F g// =mgsin  F gL =mgcos  F g =mg n=-F gL 

PHY Gravity, mass and weights. Weight=mass times gravitational acceleration F g (N)=M(kg) g(m/s 2 ) Newton’s law of universal gravitation: F gravitation =Gm 1 m 2 /r 2 G=6.67· Nm 2 /kg 2 For objects on the surface of the earth: m 1 =m earth =fixed r=“radius” of earth=fixed The earth is not a point object relative to m 2

PHY Measuring mass and weight. Given that g earth =9.81 m/s 2, g sun =274 m/s 2, g moon =1.67 m/s 2, what is the mass of a person on the sun and moon if his mass on earth is 70 kg? And what is his weight on each of the three surfaces? The mass is the same on each of the surfaces On Earth: w=686.7 N On the Moon: w=116.7 N On the Sun: N

PHY Jumping! The pelvis has a mass of 30.0 kg. What is its acceleration? Decompose all forces in x and y directions Total Force: F=  ( )=139 N Direction:  =tan -1 (F y /F x )=35.2 o Acceleration: a=F/m=139/30.0=4.65 m/s 2

PHY Tension T The magnitude of the force T acting on the crate, is the same as the tension in the rope. Spring-scale You could measure the tension by inserting a spring-scale...

PHY Newton’s second law and tension m1m1 m2m2 No friction. n FgFg T T FgFg Object 1:  F=m 1 a, so T=m 1 a Object 2:  F=m 2 a, so F g -T=m 2 a m 2 g-T=m 2 a Combine 1&2 (Tension is the same): a=m 2 g/(m 1 +m 2 ) What is the acceleration of the objects?

PHY Problem T Draw the forces: what is positive & negative??? FgFg FgFg T For 3.00 kg mass:  F=ma T-9.81  3.00=3.00  a For 5.00 kg mass:  F=ma 9.81  5.00-T=5.00  a What is the tension in the string and what will be the acceleration of the two masses? T=36.8 N a=2.45 m/s 2

PHY Friction Friction are the forces acting on an object due to interaction with the surroundings (air-friction, ground-friction etc). Two variants: Static Friction: as long as an external force (F) trying to make an object move is smaller than f s,max, the static friction f s equals F but is pointing in the opposite direction: no movement! f s,max =  s n  s =coefficient of static friction Kinetic Friction: After F has surpassed f s,max, the object starts moving but there is still friction. However, the friction will be less than f s,max ! f k =  k n  k =coefficient of kinetic friction

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PHY Problem  F g// =mgsin  F gL =mgcos  F g =mg n=-F gL  F s,k A)If  s =1.0, what is the angle  for which the block just starts to slide? B)The block starts moving. Given that  k =0.5, what is the acceleration of the block? A) Parallel direction: mgsin  -  s n=0 (  F=ma) Perpendicular direction: mgcos  -n=0 so n=mgcos  Combine: mgsin  -  s mgcos  =0  s =sin  /cos  =tan  =1 so  =45 o B) Parallel direction: mgsin(45 o )-  s mgcos(45 o )=ma (  F=ma) g(½  2-¼  2)=a so a=g¼  2

PHY All the forces come together... FgFg T n FkFk FgFg T If a=3.30 m/s 2 (the 12kg block is moving downward), what is the value of  k ? For the 7 kg block parallel to the slope: T-mgsin  -  k mgcos  =ma For the 12 kg block: Mg-T=Ma Solve for  k

PHY General strategy If not given, make a drawing of the problem. Put all the relevant forces in the drawing, object by object. Think about the axis Think about the signs Decompose the forces in direction parallel to the motion and perpendicular to it. Write down Newton’s first law for forces in the parallel direction and perpendicular direction. Solve for the unknowns. Check whether your answer makes sense.

PHY Next Lecture: Revision and go through an exam May the Force be with you!