Homework 3

About this Assignment

Due: Wed Sep 28 2005 11:29 PM EDT

Description Chapter 5 Instructions Answer narrative questions, I will update grade later.

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1. Walker2 5.P.002. [235996] In a grocery store, you push a 15.5 kg shopping cart with a force of 13.5 N. If the cart starts at rest, how far does it move in 3.00 s?
F_net = m a
Neglect Friction
F_net = 13.5 N
a_x= a = F_net / m = (13.5 N) / (15.5 kg) = 0.871 N/kg = 0.871 m/s²
x = x₀ + v_0x t + (1/2) a_x t² = 0 + 0 + ( 0.871 m/s²) (3.00 s) ² = 3.92 m
[3.92] m

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2. Walker2 5.P.009. [236001] Driving home from school one day you spot a ball rolling out into the street (Figure 5-18). You brake for 1.00 s, slowing your 880 kg car from 16.0 m/s to 9.50 m/s.

Figure 5-18

(a) What was the average force exerted on your car during braking?
a = (v_f - v_i ) / ( t_f - t_i ) = (9.50 m/s - 16.0 m/s) / (1.00 s) = -6.5 m/s²
F = m a = (880 kg) ( -6.5 m/s²) = -5720 kg m/s² = -5720 N
Magnitude
[5.72] kN
Direction

forward

backward - Correct!

(b) How far did you travel while braking?
m

Eq. 2-12
v² = v₀² + 2 a (x-x ₀)
(x-x₀)  = (1/2) (v² - v₀² ) / a = [ (9.5 m/s)² - (16.0 m/s)² ] / [ 2 (-6.5 m/s ²)]
x-x0 = (-165.8 m²/s²) / (-13.0 m/s²) = +12.75 m

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3. Walker2 5.P.015. [236003] A 65 kg parent and a 16 kg child meet at the center of an ice rink. They place their hands together and push.

(a) Is the force experienced by the parent more than, less than, or the same as the force experienced by the child?

These two forces are action-reaction partners.
---Select--- more than less than the same as
(b) Is the acceleration of the parent more than, less than, or the same as the acceleration of the child?

---Select--- more than less than the same as
Explain.
Force equals mass times acceleration. Parent and child each experience the same magnitude of force. Since the parent has a much greater mass, he/she experiences a much smaller acceleration a = F/m

(b) If the magnitude of the child's acceleration is 2.2 m/s² in magnitude, what is the magnitude of the parent's acceleration?

F = (16.0 kg)(2.2m/s²)

A(parent) = F/M(parent) = (2.2m/s²) * (16.0 kg) / 65 kg).

A(parent)= 0.541 m/s²
m/s²

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4. Walker2 5.P.016. [236004] As shown in Figure 5-19, a force of magnitude 7.50 N pushes three boxes with masses m₁ = 1.30 kg, m₂ = 3.00 kg, and m₃ = 5.20 kg.

Figure 5-19

(a) Find the contact force between boxes 1 and 2.

Treat all three boxes as one object

F = (m1+m2+m3)a

A = (7.50 N) / (1.30kg + 3.00 kg + 5.2 kg) = (7.5 N) / (9.5 kg) =0.789 m/s²

Now treat m₁ by itself

F_Net = F- Contact force = m₁ a

Contact Force = F – m₁ a = (7.50 N) – (1.30 kg)(0.789 m/s²)
N
(b) Find the contact force between between boxes 2 and 3.

Treat m3 by itself

F_Net = Contact force(2) = m₃ a

Contact Force(2) = (5.20 kg)(0.789 m/s²)
N

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5. Walker2 5.P.021. [236006] A shopper pushes a 7.3 kg shopping cart up a 13° incline, as shown in Figure 5-21. Find the horizontal force, F, needed to give the cart an acceleration of 1.63 m/s².
See Free Body Diagram and discussion below.
N

Figure 5-21

Measure all angles counter clockwize from +x-axis.
Angle of F with respect to x-axis = (360deg - 13°)
Angle of Mg with respect to x-axis = (270 deg - 13° )
x- and y- components of Newton's 2nd Law:
F_Net,x = M a_x
F_Net,y = M a_y
Bold face means vector
F_Net = F + N + Mg
a_x  = 1.63 m/s²
a_y = 0
F_x  = F cos(360-13) = F cos( 13°)
F_y = F sin(360-13) = - F sin( 13°)
N_x = 0
N_y = N
Mg_x = Mg cos(270- 13°) = - M g sin( 13°)
Mg_y = M g sin(270- 13°) = - M g cos( 13°)
x-component of Newton's 2nd Law:
F cos( 13°) + 0 + (7.3 kg )(9.81 m/s² ) cos(270- 13°) = (7.3 kg)(1.63 m/s² )
Solve for F!!
F = (7.3 kg)[ 1.63 m/s² - (9.81 m/s ²)(-0.225) ]/ cos( 13°) = 7.3 kg [ 3.84 m/s² ]/0.9744 = 28.7 N
You can solve y-component equation for N if you want, but it is not necessary.

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6. Walker2 5.P.022. [236007Two crewmen pull a boat through a lock, as shown in Figure 5-22. One crewman pulls with a force of F₁ = 120 N at an angle of = 35 ° relative to the forward direction of the boat. The second crewman, on the opposite side of the lock, pulls at an angle of 45°. With what force F₂ should the second crewman pull so that the net force of the two crewmen is in the forward direction?
- F₁ sin( 35 °) + F₂ sin(45°) = 0
F₂ = F₁ sin( 35 °) / sin(45°) = (120 N) (0.5736) / (0.7071) = 97.3 N
[97.3] N
Figure 5-22

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7. Walker2 5.P.026. [236011] An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 7.3 N; a second force has a magnitude of 4.1 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.
Vectors:
F₁ + F₂ +  F ₃ = 0
F₃ = - (F₁ + F₂)
Components
F_1x  = (7.3 N)
F_1y  = 0
F_2x  = 0
F_1y  = - 4.1 N
F_3x  = - (7.3 N) -0= -7.3 N
F_3y  = 0 - ( - 4.1 N) = +4.1 N
F₃  is in 2nd quadrant, angle from x-axis between 90 and 180 degrees.
Be careful, your calculator will not necessary place the arctangent or arccosine in the
correct quadrant.  You must adjust by hand (by adding or subtracting 180deg as neccesary).
Magnitude =  SQRT[ (-7.3 N)² + (4.1 N) ² ] = 8.37 N
Angle = arccos (F_3x / F) = arccos (-0.8719) = 150.7deg

[8.37] N
[151]° (counterclockwise from the +x axis)

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8. Walker2 5.P.032. [236013] At the bow of a ship on a stormy sea, a crewman conducts an experiment by standing on a bathroom scale. In calm waters, the scale reads 180 lb. During the storm, the crewman finds a maximum reading of 230 lb and a minimum reading of 136 lb.

(a) Find the maximum upward acceleration experienced by the crewman.
Scale force upwards = N = apparent weight
N-mg = ma
When a=0, N=mg=180 lb
m = 180lb / g
a = N/m - g = N / (180lb/g) - g = g*(N/180lb) - g = g*[ ( N/180lb) -1]
upwards acceleration when N>mg
a = (9.81  m/s²) *[(230lb)/(180 lb) -1] = 2.72 m/s ²
[2.72] m/s²
(b) Find the maximum downward acceleration experienced by the crewman.
a = (9.81  m/s²) *[(136 lb)/(180 lb) -1] = -2.40 m/s²
[2.4] m/s²

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9. Walker2 5.P.066. [236034] Seatbelts provide two main advantages in a car accident: (i) they keep you from being thrown from the car, and (ii) they reduce the force that acts on you during the collision to survivable levels. The second benefit can be illustrated by comparing the net force exerted on the driver of a car in a head-on collision with and without a seatbelt.

(a) A driver wearing a seatbelt decelerates at the same rate as the car itself. Since modern cars have a "crumple zone" built into the front of the car, the car will decelerate over a distance of roughly 1.0 m. Find the magnitude of the net force acting on a 58 kg driver who is decelerated from 18 m/s to rest in a distance of 1.0 m.

v² – v₀² = 2 a x

a = (v² – v₀² ) / (2x)

a = (0 – (18m/s)²)/(2*1.0m) = -162 m/s²

F = ma

|F| = (58 kg) (162 m/s²) = 9396 N
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(b) A driver who does not wear a seatbelt continues to move forward with a speed of 18 m/s (due to inertia) until something solid is encountered. In this case, the driver comes to rest in a much shorter distance -- perhaps only a centimeter. Find the net force acting on a 58 kg driver who is decelerated from 18 m/s to rest in 1.0 cm.

The acceleration scales inversely with distance.

Since the stopping distance is 100 times shorter,

The stopping force is 100 times greater
kN

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10. Walker2 5.P.044. [236021] An ant walks slowly away from the top of a bowling ball, as shown in Figure 5-27. If the ant starts to slip when the normal force on its feet drops below one-half its weight, at what angle does slipping begin?
°

Figure 5-27

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