CHAPTER 2 : VECTOR

VECTOR

Question 1

An 8N-force F is directed 160 degree counterclockwise from the direction of positive x-axis in the xy-plane. Sketch an arrow representing F, placing the tail of the vector at the origin. Also, draw and label the rectangular component of F on your sketch.

                                                                                - 5 m/s

                                                                         -8 m/s 

Question 2

A velocity V has rectangular component Vx = -3m/s and Vy = 5m/s. Sketch a vector diagram showing the rectangular components of V. Identify the polar component of V.

                         Magnitude =  square root of 5^2 + (-3)^2

                                           = square root of 25 + 9

                                               = square root of 34

                                               = 5.83m/s

                        Direction = Tan tehta = Vy/Vx

                                                             = 5/-3

                                                                  = - 1.667

                                                                  = Tan^-1 (-1.667)

                                                                  = -59 degree

                                                      tehta = 59 degree

Question 3

Acceleration a has rectangular component ax = -5m/s and ay = -8m/s. Sketch a vector diagram showing the rectangular components of a. Calculate the polar components (the magnitude and direction)of a.

                                   Magnitude = square root of (-5)^2 + (-8)^2

                                                          = square root of 25 + 64

                                                          = square root of 89

                                                          = 9.43 m/s^2

                                   Direction  = Tan tehta = -8/-5

                                                                               = 1.6

                                                                               = Tan ^-1 (1.6)

                                                               tehta     = 58 degree

VECTOR

1)   Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.Then,determine the direction of the hiker’s overall displacement.

ANSWER:

              

                 

     So that,the overall of resulting displacement is 15.6 km ,45 degre.

2)  The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido an angle of 40 degrees, determine the components of this force.

     

     ANSWER:

3)  A student drives his car 6.0 km, North before making a right hand turn and driving 6.0 km to the East. Finally, the student makes a left hand turn and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the student?

ANSWER:

        STEP 1                                            STEP 2                                                         

                                                    

                    R=(6.0KM)²+(8.0KM)²

                        =100 KM

                        =(100 KM)^(1/2)

                        =10 KM

1.   1)  Mac and Tosh are doing the Vector Walk Lab. Starting at the door of their physics classroom, they walk 4.0 meters, south. They make a right hand turn and walk 16.0 meters, west. They turn right again and walk 24.0 meters, north. They then turn left and walk 36.0 meters, west. What is the magnitude of their overall displacement?

                   

                 ANSWER:

 

                STEP 1                                                                                     

                                   

              STEP 2

               

R²=(52.00 m)²+(22.00m)²

R² =3188.00m

    =

   =56.50m

2.     2) A ship in the ocean of Indonesia in the lane departure from south to north with a speed of 60 km / hour. At that moment, the wind blows from west to east with a speed of 10 m/ s, so the ship led the way to the northeast. The skipper does not know the tack to the east.Determine the skipper to tack toward the east.

completion:

Given:

V_WIND (Vx) = 10 m / s

V_SHIP (Vy) = 72 km / h = 20 m / s

asked:

Tack towards the east (Θ) = ... ... ... ... ... ... ... ....?

Answer:

Tack towards the east (Θ) = arc tan (tan Θ)

= Arc tan (sin Θ / cos Θ)

= Arc tan 20/10

= Arc tan 2

= 63.4350

So tack towards the east is 63.4350

1.     3) A river flows at 3 m/s and is 300 m wide. A man swims across the river with a velocity of 2 m.s directed always perpendicular to the flow of current. Find the magnitude of the resultant velocity of the man under the effect of the stream?

Given :

·         Velocity of river  = 3 m/s

·         Width of river,  = 300 m

·         Velocity of swimmer,  = 2 m/s

·         Angle between velocity of river and velocity of swimmer,  = 90°

The following diagram illustrates the given situation:

In the above diagram, velocity ' ' is the resultant velocity of the man under the influence of the stream.

Since the resultant velocity of the swimmer ( ) is the result of the combined velocities of the river and the swimmer, therefore

By applying the vector addition formula derived from triangle law of vector addition, we get