APPLICATIONS

BASIC RESULTANT PROBLEMSFirst let’s recall some basic facts from geometry about

parallelograms

oppo

site

sid

es a

re

para

llel a

nd e

qual

opposite sides are

parallel and equal

These 2 angles are

equal

These 2 angles are

equalThese 2 angles

are equal

These 2 angles

are equal

These 2 angles add to 180°

These 2 angles add to 180°

These 2 angles add to 180°

These 2 angles add to 180°

alternate interior angles are congruent (equal sizes)

Two draft horses are pulling on a tree stump with forces of 250 pounds and 280 pounds as shown. If the angle between the forces is 60°, then what is the magnitude of the resultant force? What is the angle between the resultant and the 280 pound force?

280 lb

250

lb

Let’s take these forces as vectors and make a parallelogram and use its properties.

60°

Two draft horses are pulling on a tree stump with forces of 250 pounds and 280 pounds as shown. If the angle between the forces is 60°, then what is the magnitude of the resultant force? What is the angle between the resultant and the 280 pound force?

280 lb

250

lb

60°

v

You can make a parallelogram and the diagonal will be the resultant force showing which direction and how strong the resultant force will be.

280 lb

250

lb

60°

v

Using geometry, can you figure out any of the angles or sides of the lower triangle formed with the vector v as one side?

250

lb

120°

Opposite sides are equal and the two angles add to 180°

Looking at the lower triangle we have side, angle, side so we can use the Law of Cosines to find the magnitude (length) of v.

How can we then find as shown?

sin sin120

250

v

2 2 2280 250 2 280 250 cos120 v

459.2v

28.1

280 lb

250

lb60°

x

y

v

We could also solve this problem by making a coordinate system with the forces as vectors with initial point at the origin.

We then put the vectors in component form and add them.

280,0

250cos 60 ,250sin 60

280,0 250cos 60 ,250sin 60

2 2280 250cos 60 250sin 60 459.2 v

1 250sin 60tan 28.1

280 250cos60

INCLINED PLANE

PROBLEMS

Think of pushing or pulling something up a ramp. Our model will assume you have a well-oiled dolly and we will neglect friction. What other forces are there in this problem?

Gravity is acting on the object so the weight of the object is a force. Gravity pulls down so the force vector for the weight of the object is always vertical.

The force required to move the object is in a direction parallel to the ramp.

If we make the ramp steeper, will either of these forces change?

Gravity does NOT change. The force due to gravity stays vertical and of the same magnitude so the gravity vector remains the same.

The force required to move the object would need to be greater.

If we make the ramp even steeper what would have to change?

Let’s look at the resultant force in each case.

Can you see what happens to the resultant force as the ramp gets steeper?

Workers at the zoo must move a 250-pound giant tortoise to his new home. Find the amount of force required to pull him up a ramp leading into a truck if the angle of elevation of the ramp is 30°.

First look at this triangle and find the missing angle.

60°

Now look at this triangle. We can easily find the other angle and the hypotenuse since it is part of a parallelogram and parallel to the 250 lb. side.

30°250 lb

Okay---can you see how to use trig to find the magnitude of the force vector to pull the turtle up the ramp?

oppsin 30

hyp 250

F 250sin 30 125 lb.F

We can represent the speed and direction of a plane in still air as a vector. We’d need to add to that, the speed and direction of the wind. The resultant vector would be the speed and direction the plane would actually travel.

In these problems, they will tell you the direction the wind is coming FROM---NOT the direction it is blowing.

For example: If a wind is blowing from the southwest, it is blowing towards the northeast at a 45° angle.

from southwest

to northeast

45°

In air navigation, the bearing is a nonnegative angle smaller than 360° measured in a clockwise direction from due north.

bearing measured clockwise

from north

In this case, 270° plus the drift angle.

An airplane heads west at 350 miles per hour in a 50 mph northwest wind. Find the ground speed and bearing of the plane.

Let’s draw a picture on coordinate axes.

x

y

350 50

This is the northwest quadrant so wind would blow towards southeast.

You could draw the parallelogram and use a triangle and trig to find the resultant vector whose magnitude is the groundspeed and use the angle to determine the bearing, or you could put these 2 vectors in component form and add. Let’s do the second way this time.

v

u

350,0 v

50cos315 ,50sin 315 u

315°

c

c = v + u 350,0 50cos315 ,50sin 315

ground speed = = 316.6 mph c 1 50sin 315tan 6.4

350 50cos315

An airplane heads west at 350 miles per hour in a 50 mph northwest wind. Find the ground speed and bearing of the plane.

x

y

350 50v

u

350,0 v

50cos315 ,50sin 315 uc

c = v + u 350,0 50cos315 ,50sin 315

ground speed = = 316.6 mph c 1 50sin 315tan 6.4

350 50cos315

Remember the bearing is measured clockwise from north

5.2°

bearing = 270° – 6.4° = 263.6°