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### Two Dogs Pulling a Sled

```
Date: 10/15/2001 at 17:32:43
From: mike
Subject: Son's homework question

Two dogs are pulling a sled across an icy road. We assume that the
friction between the sled and the road is negligible. A person keeps
the sled on the proper path by pulling on it with a rope attached
at the point where the ropes from the dogs are attached.

A noise startles the dogs and they move toward opposite sides of the
road, one at an angle of 60 degrees with the centerline and the other
at an angle of 30 degrees with the centerline, each with a force of
200 N. What are the magnitude and direction of the force that the
person will have to exert on the sled to keep it moving along the
centerline of the road at constant velocity?

Mike
```

```
Date: 10/17/2001 at 10:45:19
From: Doctor Code
Subject: Re: Son's homework question

Hi Mike,

First let's draw a picture of the situation:

Y
D                         ^
/                          |
/                           |
/                            |
/  60 deg                      -------> X
\  30 deg
\
\
\
D

S is the sled, P is the person, the D's are the dogs, and the
horizontal line is the center line of the road. The angles on the
picture look the same because I can't draw them correctly on the
keyboard, but they should be 60 degrees and 30 degrees from the center
line. I have also drawn the positive X and Y axes. You can put these
axes in any orientation, but I have chosen this one so that the X axis
coincides with the direction of the road.

Notice that if both dogs were pulling with equal force at the same
angle (both at 60 degrees or both at 30 degrees), the sled would
travel along the center line. But since the angles aren't equal, the
sled is being pulled off course.

The first thing to do is to find the total force the dogs are
exerting on the sled. The forces are vectors. We can't just add the
vectors together, because they are pointing in different directions.
In order to add them, we must first resolve the vectors into X and Y
components.

For example, the 60 degree dog's vector is:

D
/
/
/
/
S

If we resolve it into X and Y components, it will look like this:

^
|
|
|
|
S---->

The above picture shows one vector along the X direction and one along
the Y direction. If we do the same thing with the other dog, we can
add the two X vectors together, and the two Y vectors. In order to
turn the vectors into component vectors, we have to use some
trigonometry:

D
/|
/ |
/  |
/60 |
S---->

The angle from the X axis to line SD is 60 degrees. We know that SD is
200 N. We need to find the bottom leg of the triangle:

X component = 200 * cos 60

The Y component is found in the same way, except you use sin instead
of cos.

Now we can add the components together:

X:

Fx = 200 * cos 60 + 200 * cos 30

= 273.2 N

Y:

Fy = 200 * sin 60 - 200 * sin 30

= 73.2 N

Note that the second force in the y direction is negative, because the
30 degree dog is on the other side of the road.

The total value of this new force is:

F = sqrt((Fx)^2 + (Fy)^2)

= sqrt(273.2^2 + 73.2^2)

= 282.8 N

And the direction of this force can be found with the tangent:

tan(angle) = 73.2/273.2

angle = arctan(73.2/273.2)

= 15 degrees

This means that the two dogs working together produce the same force
as a big dog pulling at 15 degrees above the road line with a force of
282.8 N.

Since we want the sled to travel with constant velocity along the
road, the force that the person exerts must exactly cancel the
combined force of the dogs. This will produce zero acceleration and
constant velocity. So the person must pull with 282.8 N on the other
side of the sled, but 15 degrees below the center road line:

D
/
/
/
/  60 deg
15 / \  30 deg
/   \
/     \
/       \
P         D

Write back if you need more help.

- Doctor Code, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra
High School Physics/Chemistry
High School Trigonometry

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