A child on a toboggan (combined weight 70kg) is pulled from rest on a level surface by two friends. The first friend pulls with a force of 100N in the North-West direction and the second friend pulls with a force of 60N in the direction 20 degrees East of North. What is the net force exerted on the toboggan by the two friends? In which direction will the toboggan start to move? If the acceleration of the toboggan is observed to be 0.5 metres per second squared, what is the resistive force exerted by the snow-surface on the toboggan?

First let's start with the vector addition:

If we call north the positive y-direction and east the positive x-direction, we can separate the vectors into their x- and y-components. (This is the logical choice since it's conventional to draw a compass with north pointing upward and east pointing to the right, and it's also conventional to draw a 2-dimensional plot axis with y pointing upward and x pointing to the right). For example, the first friend's force is in the NW direction, which corresponds to an angle of 135 degrees from the x-direction (a.k.a. east). Thus that vector can be written as

,

where and are unit vectors in the x- and y-direction (a.k.a. east and north), respectively.

Next, you'd repeat the same process for the force from the second friend. Once you have both forces broken into their x- and y-components, you can just add like components together to get the net force. You can calculate its magnitude as

,

where Fx and Fy are the total forces in the x- and y-directions respectively.

You can calculate its direction as



A word of warning here: when calculating the arctan, your calculator or computer will probably give you the closest angle within +/-90 degrees. However, bear in mind that any angle that's an integer multiple of 180 degrees more or less than the answer your calculator gives is also a possibility. So you'll need to use some common sense when choosing which angle corresponds to the actual direction. For example, let's say your calculator computes the angle at around -73 degrees. That means 73 degrees clockwise from east, which is in the southerly direction (roughly SSE). You know that both friends are pulling in a northerly direction (one to the NW and one 20 degrees east of north), so common sense tells you that -73 degrees is not the correct direction. The correct answer must be 180 degrees away, in the completely opposite direction, at 107 degrees.

Finally, what must be the total force on the 70-kg toboggan if it's accelerating at 0.5 meters per second squared? You already calculated the magnitude of the net force with which the friends are pulling, so find the difference between the two, and that's got to be the frictional force from the snow.

The other person push the toboggan with force 1 newton form north-west direction. By applying the vector addition and Newtons law.

Uhhh... what?

That comment, satender, was very random :eek:

Many Thanks. Your answer makes understanding Vector Addition a whole lot easier.
Just to re-cap, I used the first equation which you provided to work out F1, which gave me:
(-70.71x + 70.71y) Newtons. Then I worked out F2, which gave me (+20.52x + 56.38y) Newtons. Working out F(net) from the second equation using these values then gave me 136.64 Newtons.
To calculate the direction, I then used the third equation which you provided: F(net) = tan (-1) (127.09 / (-50.19)) = -68.45 degrees. Therefore, the correct answer, as you pointed out, should be 180 degrees away, at 111.45 degrees from the x axis (which is pointing to the right). I think this then would be 17 degrees West of North.
Finally, working out the resistive force, I used Newton's 2nd Law: F(net) - F(k) = ma. Therefore, F(k) = F(net) - ma = 136.64 Newtons - (70)(0.5) Newtons = 101.64 Newtons. I'm hoping I did this right.

Well done! Everything looks good except for one little thing: you calculated the degrees West of North wrong (or, more likely, you just typed it wrong when recapping). North is 90 degrees from the x-axis. So 111.45 degrees would be 111.45 - 90 = 21.45 degrees West of North, not 17. Otherwise, I think everything else is correct.

Sorry, my mistake. Thanks again for solving the problem.