Find the magnitude of the resultant of forces represented by 3i – 2j + 3k, i – 3j and 2i + 4k . +1 more q.?
3. Find, in the form , the resultant of three horizontal forces with magnitudes and directions of 16 N in the direction N60°W, 8 N in direction N30°E and 11 N in the direction SE.
can anyone help me with these? for the first one i got 10.5, is that right? Thanks
Update:Sorry lumax it says in the form aJ+bJ
More Questions From This User See All
Answers & Comments
Verified answer
For the first part:
(3i - 2j + 3k) + (i - 3j) + (2i + 4k) = 6i - 5j + 7k
The magnitude is sqrt(6^2 + 5^2 + 7^2) = sqrt(110). I wouldn't round that off to 10.5--use the exact value unless the instructions say otherwise.
The second part only asks about horizontal forces. All you have to do is multiply the force by the sine of the angle that it deviates from due north or due south. (For example, a force due north has sin(0) so no horizontal force is applied, whereas a force due east has sin(90°) so the entire force is horizontal.) If east is positive and west is negative:
-16 N sin(60°) + 8 N sin(30°) + 11 N sin(45°)
-8 sqrt(3) N + 4 N + 11 N / sqrt(2)
The exact answer is that you have a horizontal force of (-8 sqrt(3) + 4 + 11 sqrt(2) / 2) N, but in this case the problem probably wants you to approximate, so the answer would be about -2.0782 N, or 2.0782 N to the west. Since this is presented in terms of a physics problem, you may need to round to a lower number of significant digits. At two signficiant figures, that's 2.1 N to the west.