7. Four perpendicular forces, 2 N, 3 N, 9 N and 10 N act on object P as shown in Diagram 2. 10 N A B 9 N 3 N 2 N P Diagram 2 Wha
t is the magnitude of the resultant force? 10 N C 18 N 14 N D 20 N
1 answer:
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Answer:
(i) See attached image for the drawing
(ii) net force given in component form: (20, 20)N with magnitude:
Explanation:
First try to write all forces in vector component form:
The force F1 acting at 45 degrees would have multiplication factors of on both axes, to take care of the sine and cosine projections. Therefore, the:
x-component of F1 is
y-component of F1 is
As far as force F2, it is given already in x and y components, then:
x-component of F2 = 8 N
y-component of F2 = -6 N (negative meaning pointing down the y-axis)
Force F3 has only component (upwards) in the y-direction
x-component of F3 = 0 N
y-component of F3 =14 N
The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:
x-component of R = 12 + 8 = 20 N
y-component of R = 12 + 14 - 6 = 20 N
Therefore, the acceleration that the mass receives due to this force is given in component form as:
x-component of acceleration: 20 N / 5 kg =
y-component of acceleration: 20 N / 5 kg =
Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:
x-component of the velocity is:
y-component of the velocity is:
