Vectors - Physics - Solved Homework, Exercises of Physics

Solved assignment for chapter "Vectors"

Typology: Exercises

2015/2016

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SOLUTION TO
HOMEWORK PROBLEMS Chapter 3: VECTORS
1. A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin
of a two-dimensional Cartesian coordinate system. If the fly is located at the point having
coordinates (3.00, 1.50) m, what is its location in polar coordinates?
θ = tan-1 (y/x) > θ = tan-1(1.5/3.0) > θ = 26.6
x = r Cos θ > r = 3.00 / cos 26.6 > r = 3.35
2. A surveyor measures the distance across a straight river by the following method. Starting
directly across from a tree on the opposite bank, she walks d = 138 m along the riverbank to
establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is
= 35.0°. How wide is the river?
Tan 35 = y / d > y = 0.7 * 138 = 96.6 m
3. A force
1
F
of magnitude 6.00 units acts on an object at the origin in a direction
= 30.0°
above the positive x axis. A second force
2
F
of magnitude 5.00 units acts on the object in the
direction of the positive x axis. Find magnitude and direction of the resultant force
12
FF
6Cos30i + 5i + 6sin30j = (5.2+5)i + 3j > mag = √(10.22 + 32) = 10.6 N
θ = tan-1 (3/10.2) = 16.4
4. A person walks 25.0° north of east for 3.10 km. How far would she have to walk due south
and due west to return to the starting point?
W = 3.10 Cos 25 = 2.81 km
S = 3.10 Sin 25 = 1.31 km
5. A map suggests that Atlanta is 730 miles in a direction of 5.00° north of east from Dallas.
The same map shows that Chicago is 560 miles in a direction of 21.0° west of north from Atlanta.
(Figure P3.22 shows the locations of these three cities.) Modeling the Earth as flat, use this
information to find the displacement from Dallas to Chicago.
DC east DA east AC east
DC nort h DA north AC nort h
+ 730cos 5.00 560sin 21.0 527 miles
+ 730sin 5.00 +560cos 21.0 586 miles
d d d
d d d
By the Pythagorean theorem,
22
DC east DC north
( ) ( ) 788 mid d d
.
Then
DC north
DC east
tan 1.11
d
d

and
48.0

.
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SOLUTION TO

HOMEWORK PROBLEMS Chapter 3: VECTORS

1. A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system. If the fly is located at the point having coordinates (3.00, 1.50) m, what is its location in polar coordinates?

θ = tan-1^ (y/x) > θ = tan-1(1.5/3.0) > θ = 26.

x = r Cos θ > r = 3.00 / cos 26.6 > r = 3.

2. A surveyor measures the distance across a straight river by the following method. Starting directly across from a tree on the opposite bank, she walks d = 138 m along the riverbank to establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is  = 35.0°. How wide is the river?

Tan 35 = y / d > y = 0.7 * 138 = 96.6 m

3. A force F 1 of magnitude 6.00 units acts on an object at the origin in a direction  = 30.0°

above the positive x axis. A second force F 2 of magnitude 5.00 units acts on the object in the

direction of the positive x axis. Find magnitude and direction of the resultant force F 1  F 2

6Cos30i + 5i + 6sin30j = (5.2+5)i + 3j > mag = √(10.2^2 + 3^2 ) = 10.6 N

θ = tan-1^ (3/10.2) = 16.

4. A person walks 25.0° north of east for 3.10 km. How far would she have to walk due south and due west to return to the starting point?

W = 3.10 Cos 25 = 2.81 km

S = 3.10 Sin 25 = 1.31 km

5. A map suggests that Atlanta is 730 miles in a direction of 5.00° north of east from Dallas. The same map shows that Chicago is 560 miles in a direction of 21.0° west of north from Atlanta. (Figure P3.22 shows the locations of these three cities.) Modeling the Earth as flat, use this information to find the displacement from Dallas to Chicago.

DC east DA east AC east DC north DA north AC north

  • 730 cos 5.00 560 sin 21.0 527 miles
  • 730 sin 5.00 +560 cos 21.0 586 miles

d d d d d d

By the Pythagorean theorem, d  ( d DC east )^2  ( d DC north)^2 788 mi.

Then DC north

DC east

tan 1.

d d

   and   48.0.

6. An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 800 m, horizontal distance 19.2 km, and 25.0° south of west. The second aircraft is at altitude 1 100 m, horizontal distance 17.6 km, and 20.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.)

Position vector of A = 19.2 Cos (-25) i + 19.2 Sin(-25) j + 0.8 k = 17.4 i - 8.11 j +0.8 k

Position vector of B = 17.6 Cos (-20) i + 17.6 Sin(-20) j +1.1 k =16.5 i - 6.0 j + 1.1 k

Relative position = A – B

= 0.9 i – 2.1 j - 0.3 k km

7. The three forces shown act on a particle. What is the direction of the resultant of these three forces?

R = (65 cos 30 + 30 Cos (180) + 20 Cos (250)) i + (65 sin 30 + 30 sin (180) + 20 sin (250)) j

= 19.45 i + 13.7 j

Mag = (19.45^2 + 13.7^2 )0.5^ = 23.8 N

8. A person going for a walk follows the path shown in Figure P3.47. The total trip consists of four straight-line paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?

Disp = (100 + 0 + 150 cos 210 + 200 Cos 120) i + ( 0 – 300 + 150 sin 210 + 200 sin 120) j = -129.9 i - 201.8 j m