Subtraction of Vectors, and Multiplication of a Vector by a Scalar
Adding Vectors by Components
Solving Problems Involving Projectile Motion
3-1 Vectors and Scalars A vector has magnitude as well as direction . Some vector quantities: displacement , velocity , force , momentum A scalar has only a magnitude . Some scalar quantities: mass , time , temperature
3-2 Addition of Vectors – Graphical Methods For vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs , as the figure indicates. 8 km + 6 km = 14 km East 8 km - 6 km = 2 km East
3-2 Addition of Vectors – Graphical Methods If the motion is in two dimensions, the situation is somewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem .
3-2 Addition of Vectors – Graphical Methods Adding the vectors in the opposite order gives the same result:
3-2 Addition of Vectors – Graphical Methods
Even if the vectors are not at right angles, they can be added graphically by using the “ tail-to-tip ” method:
1. Draw V1 & V2 to scale.
2. Place tail of V2 at tip of V1
3. Draw arrow from tail of V1 to tip of V2
This arrow is the resultant V (measure length and the angle it makes with the x -axis )
Same for any number of vectors involved.
3-2 Addition of Vectors – Graphical Methods The parallelogram method may also be used;
1. Draw V1 & V2 to scale from common origin.
2. Construct parallelogram using V1 & V2 as 2 of the 4 sides.
Resultant V = diagonal of parallelogram from common origin (measure length and the angle it makes with the x -axis)
3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction. Then we add the negative vector:
3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar A vector V can be multiplied by a scalar c ; the result is a vector c V that has the same direction but a magnitude c V . If c is negative , the resultant vector points in the opposite direction.
3-4 Adding Vectors by Components Any vector can be expressed as the sum of two other vectors, which are called its components . Usually the other vectors are chosen so that they are perpendicular to each other.
3-4 Adding Vectors by Components If the components are perpendicular , they can be found using trigonometric functions.
3-4 Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically :
3-4 Adding Vectors by Components
Draw a diagram ; add the vectors graphically.
Choose x and y axes.
Resolve each vector into x and y components .
Calculate each component using sines and cosines.
Add the components in each direction.
To find the length and direction of the vector, use:
V = displacement 500 m, 30º N of E
Note that since both components are negative, the vector is in the 3 rd quadrant. Problem 3-14
3-5 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity ; its path is a parabola . Of the form A parabola in the x-y plane !!
It can be understood by analyzing the horizontal and vertical motions separately. 3-5 Projectile Motion
3-5 Projectile Motion The speed in the x -direction is constant; in the y -direction the object moves with constant acceleration g . This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x -direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly. Show clip
3-5 Projectile Motion If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.
3-6 Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.
3-6 Solving Problems Involving Projectile Motion
Read the problem carefully, and choose the object(s) you are going to analyze.
Draw a diagram.
Choose an origin and a coordinate system.
Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g .
Examine the x and y motions separately.
3-6 Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that v x never changes, and that v y = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.
Simplest example : Ball rolls across table, to the edge & falls off edge to floor. Leaves table at time t = 0 . Analyze y part of motion & x part of motion separately .
y part of motion : Down is positive & origin is at table top:
y 0 = 0 . Initially, no y component of velocity: v y0 = 0
v y = gt and y = ( ½) g t 2
x part of motion : Origin is at table top: x 0 = 0 . No x component of acceleration(!) : a = 0 . Initially x component of velocity is: v x0
v x = v x0 and x = v x0 t
Ball Rolls Across Table & Falls Off
Vector velocity v has
2 components :
v x = v x0 , v y = gt
Vector displacement D has
2 components :
x = v x0 t , y = (½)g t 2
General Case : Take y positive upward & origin at the point where it is shot: x 0 = y 0 = 0
v x0 = v 0 cos θ 0 v y0 = v 0 sin θ 0
NO ACCELERATION IN THE x DIRECTION !
v x = v x0 x = v x0 t
Vertical motion :
v y = v y0 - gt y = v y0 t - ( ½) g t 2
(v y ) 2 = (v y0 ) 2 - 2gy
If y is positive downward, the - signs become + signs.
a x = 0, a y = -g = -9.8 m/s 2
Range ( R ) of projectile Maximum horizontal distance before returning to ground. Derive a formula for R .
Range R the x where y = 0 !
Use v x = v x0 , x = v x0 t , v y = v y0 - gt
y = v y0 t – ( ½) g t 2 , (v y ) 2 = (v y0 ) 2 - 2gy
First , find the time t when y = 0
0 = v y0 t - ( ½) g t 2
t = 0 (of course!) and t = (2v y0 )/g
Put this t in the x formula: x = v x0 (2v y0 )/g R
R = 2(v x0 v y0 )/g, v x0 = v 0 cos( θ 0 ), v y0 = v 0 sin( θ 0 )
R = (v 0 ) 2 [2 sin( θ 0 )cos( θ 0 )]/g
R = (v 0 ) 2 sin(2 θ 0 )/g (by a trig identity)
Example 3-4: Driving off a cliff!!
How fast must the motorcycle leave the cliff to land at
x = 90 m, y = -50 m ? v x0 = ?
y is positive upward y 0 = 0 at top v y0 = 0 v x = v x0 = ? v y = -gt x = v x0 t y = - ( ½)gt 2 Time to Bottom: t = √ 2y/(-g) = 3.19 s v x0 = (x/t) = 28.2 m/s
Example 3-9, A punt!
v 0 = 20 m/s, θ 0 = 37º
v x0 = v 0 cos( θ 0 ) = 16 m/s
v y0 = v 0 sin( θ 0 ) = 12 m/s
y = y 0 + v y0 t – ( ½) g t 2
x = v x0 t
Summary of Chapter 3
A quantity with magnitude and direction is a vector.
A quantity with magnitude but no direction is a scalar.
Vector addition can be done either graphically or using components.
The sum is called the resultant vector.
Projectile motion is the motion of an object near the Earth’s surface under the influence of gravity.