Chapter 6 Newton’s Laws ‐ Examples Chapter 6 Newton’s Laws ‐ Examples Example 1 A traffic light is suspended from two massless wires AB and BC, as shown. The tensions in wires AB and BC are unknown. m = 20 kg a. Without doing calculations, which tension is greater, the tension in wire AB or BC? b. Draw a FBD of the traffic light. Include a coordinate system. Do NOT use the double subscript notation in this
problem.
c. Set up Newton’s second law equations. Do not solve the equations here.

Fx=
=
Fy=
=
Chapter 6 Newton’s Laws ‐ Examples d. Use your equations to find the tensions in wires AB and BC.
e. If you double the mass of the traffic light the tension in wire AB will increase / decrease / stay the same. If you increase the 30 degree angle to 40 degrees, the tension in wire AB will increase / decrease / stay the same. If you increase the 30 degree angle to 40 degrees, the tension in wire BC will increase / decrease / stay the same. Chapter 6 Newton’s Laws ‐ Examples Example 2 A 100 kg block with a weight of 1000N (g=10m/s2) hangs on a rope. Scenario A: The block is stationary. Scenario B: The block is moves upward with a steady speed of 5 m/s. Scenario C: The block is moving down with 15 m/s and slowing down at a rate of 5 m/s2. m
a. Draw a FBD of the block for each scenario. Include a coordinate system. Use the double subscript notation in
this problem.
Scenario B
Scenario A
Scenario C
b. Set up Newton’s second law equations for each scenario and solve for the tension.
Scenario B
Scenario A
Fy=
=
Fy=
Scenario C
=
Fy=
=

c. What is the situation in scenario B called?
d. What would happen to the tension in scenario B if the speed were increased?
The tension would increase / decrease / stay the same.
Chapter 6 Newton’s Laws ‐ Examples Example 3a A skier of mass m starts down a 50‐m long, 10° slope. There is no friction between the skier and the slope. m = 75 kg 50 m
10 ° You want to find the speed of the skier at the bottom of the slope. a. Draw a FBD of the skier. Include a coordinate system. Use the double subscript notation in this problem.
FBD
b. Set up Newton’s second law equations.

Fx=
=
Fy=
=
c. Use your equations to find the speed at the bottom.
Chapter 6 Newton’s Laws ‐ Examples Example 3b A skier of mass m starts down a 50‐m long, 10° slope. There is friction between the skier and the slope. m = 75 kg µs = 0.12 µk=0.06 50 m
10 ° You want to find the speed of the skier at the bottom of the slope. a. Draw a FBD of the skier. Include a coordinate system. Use the double subscript notation in this problem.
FBD
b. Set up Newton’s second law equations.

Fx=
=
Fy=
=
c. Use your equations to find the speed at the bottom.
Chapter 6 Newton’s Laws ‐ Examples Example 4: Burglars are trying to haul a 1000‐kg safe up a ramp to their getaway truck. The coefficients of friction between the safe and the ramp are µs = 0.8 and µk=0.6. For what value T of the tension does the safe start moving? If that value of T is still applied while the safe is moving, what is the acceleration of the safe?   Take  = 20°. Example 5: A car moving with an initial speed of 20 m/s comes to a stop over a distance 50 m. Assume that the deceleration of the car is constant. A 40‐kg passenger in the car is not wearing a seat belt! The coefficients of static and kinetic friction between the passenger and the seat are µs = 0.4 and µk = 0.3. Does the passenger slide off the seat while the car is slowing down? Legal notice: this is a hypothetical example. Don't try this outside of class! Example 6 (at home): A 2kg wood block slides down a vertical wood wall while you push on it at a 45° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed? Example 7 (at home): A 50kg student gets in a 1000kg elevator at rest. As the elevator begins to move, she has an apparent weight of 600N for the first 3s. How far has the elevator moved, and in which direction, at the end of 3s? Chapter 6 Newton’s Laws ‐ Examples Mass‐Weight‐Apparent Weight 1. Object in free fall 2. How do you feel weight? 3. How do you measure weight? 4. What if you accelerate up at 1m/s2? Chapter 6 Newton’s Laws ‐ Examples Example 8 (Drag): In many situations, drag can be modeled as ,
, where  is the density of the air (1.2 kg/m3 near the surface), C is the drag coefficient (see table), A is the cross‐sectional area facing the wind, and v is the speed. a. Jump out of helicopter, closed parachute You jump out of a stationary helicopter, you do NOT use your parachute. What happens? b. Open your parachute while falling You are falling toward Earth at terminal speed. Your life passes in front of your inner eye and you decide that it might be time to open your parachute. What happens when you do? Example 9: coffee filters A single coffee filter is held next to three stacked coffee filters. Both the single filter and the stack are dropped? a. Which will reach the ground first? b. Which will have the greater terminal speed? Example 10: Two blocks are connected via a string going over a pulley (massless, frictionless), as shown. The table has no friction and the blocks are initially at rest. The weight of block B is 500 N. The tension in the string is a. > 500 N b. < 500 N c. = 500 N Chapter 6 Newton’s Laws ‐ Examples Static Friction, : Direction: parallel to surface Magnitude ranges between zero and a maximum value: 0
‐
‐
‐
depends on normal force between the touching surfaces, and , the coefficient of static friction. Does not depends on size of contact area. Kinetic friction, : Direction: parallel to surface Magnitude does not depend on speed (within reasonable limits): ‐
‐
‐
‐
depends on normal force between the touching surfaces, and , the coefficient of kinetic friction. Does not depends on size of contact area. Does not depends on speed of object. Typically, Rolling friction, : depends on normal force between the touching surfaces, and , the coefficient of rolling friction. Think of similar to kinetic friction.