A model boat floating on a river has a force "A" from its engine of magnitude 40 Newtons acting to push it in a direction towards 30 degrees to the North of East. At the same time, the river current acts on the boat in a direction towards 45 degrees to the South of East with an effective force of magnitude 10 Newtons. Using vectors, calculate the magnitude and direction of the resultant force on the boat.
I've tried to do this using two methods, but the answers don't add up.
First, resolving horizontally
Fx = 40 Cos 30 + 10 Cos 45 = 41.71
Fy = 40 Sin 30 - 10 Sin 45 = 12.93
Therefore resultant = sqrt(41.71^2 + 12.92^2) = 43.66 N
Direction = ArcTan (12.9/41.71) = 17.18 degrees North of East
I'm fairly confident that the above working is correct.
Attempt 2,
Using triangle of forces, I've drawn a triange ABC with AB = 40, AC = 10 and BAC = 75 degrees.
i.e. two sides of the triangle are 40 and 10, with the angle inbetween these two sides being (30+45) 75 degrees.
Wouldn't the other side be the resultant.
Using the cosine rule I get
R^2 = 40^2+10^2 - 10 * 40 Cos (75)
R = 38.64 ( not what I got before i.e. 43.65)
Also, after using the sine rule, I get the angle of the resultant as 48 degrees North of East.
Which is correct (I'm guessing the first) and where have I gone wrong...
Thanks.
I've tried to do this using two methods, but the answers don't add up.
First, resolving horizontally
Fx = 40 Cos 30 + 10 Cos 45 = 41.71
Fy = 40 Sin 30 - 10 Sin 45 = 12.93
Therefore resultant = sqrt(41.71^2 + 12.92^2) = 43.66 N
Direction = ArcTan (12.9/41.71) = 17.18 degrees North of East
I'm fairly confident that the above working is correct.
Attempt 2,
Using triangle of forces, I've drawn a triange ABC with AB = 40, AC = 10 and BAC = 75 degrees.
i.e. two sides of the triangle are 40 and 10, with the angle inbetween these two sides being (30+45) 75 degrees.
Wouldn't the other side be the resultant.
Using the cosine rule I get
R^2 = 40^2+10^2 - 10 * 40 Cos (75)
R = 38.64 ( not what I got before i.e. 43.65)
Also, after using the sine rule, I get the angle of the resultant as 48 degrees North of East.
Which is correct (I'm guessing the first) and where have I gone wrong...
Thanks.
-
Hello
The first solution is correct. The resulting vector is the sum of both velocity vectors.
The resulting vector of the second solution is the difference vector of the two velocity vectors.
You can find the coordinates of this vector also as
AB - AC = (40*cos30 , 40*sin30) - ( 10*cos45, - 10*sin45)
= (40cos30 - 10cos45, 40sin30 + 10sin45) = (27,57, 27,07)
Magnitude = √(27,57^2+27,07^2) = 38,63 N, direction 44,47° (check your direction).
--------
You can also explain it so:
You form a parallelogram of forces from the two force vectors:
your first, correct, solution is the diagonal from the origin to the distant vertex,
the other, wrong, solution is the other diagonal between the two adjacent vertices.
Regards
The first solution is correct. The resulting vector is the sum of both velocity vectors.
The resulting vector of the second solution is the difference vector of the two velocity vectors.
You can find the coordinates of this vector also as
AB - AC = (40*cos30 , 40*sin30) - ( 10*cos45, - 10*sin45)
= (40cos30 - 10cos45, 40sin30 + 10sin45) = (27,57, 27,07)
Magnitude = √(27,57^2+27,07^2) = 38,63 N, direction 44,47° (check your direction).
--------
You can also explain it so:
You form a parallelogram of forces from the two force vectors:
your first, correct, solution is the diagonal from the origin to the distant vertex,
the other, wrong, solution is the other diagonal between the two adjacent vertices.
Regards