Forces in One Dimension Chapter 4 Physics Principles and Problems Zitzewitz, Elliot, Haase, Harper, Herzog, Nelson, Nelson, Schuler and Zorn McGraw Hill,

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!