Vectors All vector quantities have magnitude (size) and direction. Vectors in physics that we will use:  Force, displacement, velocity, and acceleration. A vector quantity is graphically represented by a vector.

An example is a force vector.   The label for the vector indicates the type of vector. Every vector has 4 parts:  is the point of application sometimes referred to as the point of concurrency.  F is the label symbolizing the type of vector. F

 The arrowhead indicates the direction of the vector.  The length of the vector indicates the magnitude of the vector.

Free-Body Diagrams (Force Diagrams) A free-body diagram sometimes called a force diagram shows all the forces that are exerted on a single object.  Usually a point is used to represent the object.

Examples of force diagrams: Consistency is important!  F w represents the weight of an object.  T represents the tension in a rope or string and F N the normal (perpendicular) force. FwFw T FwFw T FwFw FNFN

Guidelines for Free-Body Diagrams Only forces are to appear in diagrams. All the forces exerted on the object but none of those objects exerting those forces are included. Each force is represented by a force vector.  The tail end of the arrow is placed at the point of application of the force.  The direction of the arrow indicates the direction of the force.

Forces in Free-Body Diagrams The following is a list of common forces found in force diagrams. Normal Force (F N )  For each surface in contact with an object, there is a normal force which is perpendicular to the surfaces.

Frictional Force (F f )  For each surface in contact with an object, there is a normal force which is perpendicular to the surfaces. When a surface is described as frictionless, then F f = 0 and is not included in the diagram.

Gravitational Force (F w )  Be careful! Even when an object is said to be in a state of weightlessness, it still has weight. Tension (T or F T )  For each string or rope, there is a tension. If the string or rope is slack, then T = 0.

Air Resistance (F R )  Occurs when an object moves relative to the air, i.e. a parachute.  If an object is a small dense object moving very quickly, this is usually ignored. Net Force  The net or resultant force is the vector sum of all the forces acting on the object.

 The net or resultant force is the vector sum of all the forces acting on the object.  The net force is never drawn in the diagram!  It is the net force that determines the motion of an object.

More New Vocabulary Forces are measured in units of newtons (N).  In dynamics, specifically Newton’s 2 nd law, we will more formally define a newton. Right now, consider 1 N ≈ 1/5 pound. A resultant force is the one force that combines the effects of all the forces acting on an object.

The direction in which a force acts is sometimes referred to as bearing. Sometimes instead of using N, E, S, and W to designate the direction, an angular measure is given. 000° 090° 180° 270° You start at 000° and go clockwise.

A Typical Force Problem Two soccer players kick a soccer ball at exactly the same time. One player exerts a force of 55 N north and the second player exerts a force of 75 N east. Mathematically determine the magnitude and direction of the resultant. F1F1 F2F2 FRFR θ F1F1 F2F2 FRFR θ

The force diagram can be drawn using the parallelogram method (left) or the triangle method (right). They give the same results and sometimes one may be favored more than the other depending on the application. To solve the problem mathematically, we use the pythagorean theorem, c 2 = a 2 + b 2, which must be written in terms of forces.

To determine the direction (bearing), we use one of the trigonometric functions which must be written in terms forces. F R = (F F 2 2 ) 1/2 F R = ((55 N) 2 + (75 N) 2 ) 1/2 = 93 N θ = sin -1 F R = 93 N, 36° N of E F1F1 FRFR =sin N 93 N = 36°

Graphical Solution We will not solve the problem graphically but rather just outline the steps.  Choose a scale of ? cm = ? N, where the ?’s are chosen to conveniently draw the force (vector) diagram.  The resultant vector can be measured in cm from using either diagram.

 The resultant vector must have the same point of application as the other vectors.  According to your scale, you convert the number of cm to the number of N (if drawing a force diagram).  The angle is measured with a protractor.

Velocity Vectors A motorboat travels at 7.5 m/s W. The motorboat heads straight across a river 120 m wide. (a)If the river flows downstream at a rate of 3.2 m/s S, determine the resultant velocity of the boat. θ VBVB VrVr VRVR VrVr = 3.2 m/s S VBVB = 7.5 m/s W

V R = (V B 2 + V r 2 ) 1/2 V R = ((7.5 m/s) 2 + (3.2 m/s) 2 ) 1/2 = 8.2 m/s θ = sin -1 V R = 8.2 m/s, 23° S of W or 8.2 m/s, 247° VrVr VRVR == sin m/s 8.2 m/s 23°

(b) How long does it take the boat to reach the opposite side of the river? Δs = 120 mV B = 7.5 m/s W ΔsΔs ΔtΔt v ave = Δt = 120 m 7.5 m/s = 16 s

(c) How far downstream is the boat when it reaches the other side of the river? v ave = ΔsΔs ΔtΔt ΔsΔs = 3.2 m/s × 16 s = 51 m

The Resultant and Equilibrant One force of 17 N at a bearing of 090° acts concurrently with a force of 11 N at a bearing of 180°. (a)Determine the magnitude and bearing of the resultant. F2F2 = 11 N F1F1 = 17 N F1F1 F2F2 FRFR θ

F R = (F F 2 2 ) 1/2 F R = ((17 N) 2 + (11 N 2 ) 1/2 = 20.0 N θ = cos -1 F R = 20.0 N, 122° F1F1 FRFR == cos N 20. N 32°

(b) Determine the magnitude and bearing of the equilibrant. The equilibrant is equal in magnitude and opposite in direction (180°) to the resultant. F net = F R + F eq = 0 N F1F1 F2F2 FRFR θ F eq = 20.0 N, 302°

Resolution of Forces Harry pushes a lawn mower along a level ground. The handle of the lawn mower makes an angle of 20.0° with the ground as Harry pushes with a force of 78.0 N. (a)Determine the useful component of the force. θ FHFH FVFV FRFR θ= 20.0° FRFR = 78.0 N

. cos θ = FHFH FRFR × F H = F R cos θ = 78.0 N × = 73.3 N (b) Determine the wasted component of the force. sin θ = FVFV FRFR F V = F R sin θ = 78.0 N × = 26.7 N ×

The force diagram shows the lawn mower being pushed from right to left. It could also be shown being pushed from left to right.  The results would be the same.  If a particular direction was given, it would have to appear that way in the problem. The wasted component would be F V because that component of the force is trying to push the lawn mower straight into the ground.

Wrap Up Questions Can the magnitude of a vector have a negative value? Justify your answer. Remember that vectors are measurable quantities that have a direction as well as a magnitude. Such quantities as forces, displacements, velocities, and accelerations can not have a magnitude below zero but can point in a negative direction.

Two forces, F 1 = 12.0 N and F 2 = 4.0 N are concurrent (acting on the same point). (a)What is the minimum resultant? The minimum resultant would be 8.0 N if the angle between them was 180°. (b) What is the maximum resultant? The maximum resultant would be 16.0 N if the angle between them was 0°.