Forces in One Dimension
Chapter 4Physics Principles and ProblemsZitzewitz, Elliot, Haase, Harper,
Herzog, Nelson, Nelson, Schuler and Zorn
McGraw Hill, 2005
Remember! When using F = ma it is the net force!• Kamaria is learning how to ice skate. She wants here mother to
pull her along so that she has an acceleration of 0.80m/s2. If Kamaria’s mass is 27.2kg, with what force does her mother have to pull her along (neglect ice resistance)?
1. Fnet = Fmother on Kamaria + ( - Fkanaria on mother)
2. a = Fnet / m
4. (a • m) - ( - Fkanaria on mother) = Fmother on Kamaria
3. a = Fmother on Kamaria + ( - Fkanaria on mother) / m
5. Fmother on Kamaria = (0.80m/s2 • 27.2kg) + 0 N
Applying Newton’s Laws• The mass of an object is a magnitude of that object’s
amount of matter. It never changes (i.e. 12g of iron on earth is still 12g of iron on the moon).
• The weight of a object is a force (hence, weight’s units are Newtons) and is directly related to the mass (kg) and acceleration (m/s2) of the object.
What is your weight on earth?
What is your weight on the moon?
The Elevator Problem
Does your weight change in an elevator?
http://plus.maths.org/issue38/features/livio/figure8.jpg
Understanding Weight
• Apparent Weight - the force an object experiences as a result of ALL the forces acting upon it, resulting in acceleration.
• Weightlessness - when an object’s apparent weight is zero as a result of no contact forces being exerted on the object (actual weight is not zero however).
Real and Apparent Weight
• Your mass is 75.0kg and you are standing on a scale in an elevator. Starting from rest the elevator accelerates upward at 2.0m/s2 for 2.0s and then continues at a constant speed. What is the reading of the scale during rest and during acceleration?
http://sol.sci.uop.edu/~jfalward/physics17/chapter4/elevator.jpg
Solution• F = ma• Fnet = Fscale + ( -Fg)• Fscale = Fnet + Fg
At RestFscale = 0 + Fg
Fscale = mg
Fscale = (75kg)(10m/s2)
Fscale = 750 N
AcceleratingFscale = Fnet + Fg
Fscale = ma + mg
Fscale = (2.0m/s2)(75kg) + (75kg)(10m/s2)
Fscale = 900 N
Drag Force • The force exerted by a fluid on
the object moving through that fluid. This force is dependent
upon the properties of both the object (shape, mass) and fluid.
http://www.swe.org/iac/images/prafoil.jpg
http://www.fluent.com/about/news/newsletters/02v11i1/img/a9i5_lg.gif
http://newsimg.bbc.co.uk/media/images/42381000/jpg/_42381388_swimmers416.jpg
Terminal Velocity - the constant velocity that an object obtains when the drag force equals the
force of gravity (acceleration = 0).
http://www.iop.org/activity/education/Teaching_Resources/Teaching%20Advanced%20Physics/Mechanics/Images%20200/img_mid_4140.gif
Newton’s Third Law• Forces come in an
interaction pair (two forces that are in opposite direction and have equal magnitudes).
• FA on B = - FB on A • The force of A on B is
equal in magnitude and opposite direction of the force of B on A.
• Action and Reaction. http://www.primidi.com/images/newton_action_reaction_law.jpg
Tension - the force exerted by a string
or rope.
• A 50kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3m, it is moving at 3m/s. If the acceleration is constant, is the rope in danger of breaking?
http://upload.wikimedia.org/wikipedia/commons/8/8f/Helicopter_at_Yellowstone_1988.jpg
Solution• Fnet = Ftension + ( -Fg)
• Ftension = Fnet + Fg
• Ftension = ma + mg
Use vf2 = vi
2 + 2ad to solve for a.
since vi2 is 0 then a = vf
2 / 2d
Therefore,1. FT = (m)(vf
2 / 2d) + (m)(g)
2. FT = (50kg)((3m/s)2/2(3m) + (50kg)(10m/s2)
3. FT = 570 N, Yes it will break!