The figure shows an overhead view of a 0.028 kg lemon half and two of the three

Question

The figure shows an overhead view of a 0.028 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force Upper F Overscript right-arrow EndScripts Subscript 1 has a magnitude of 3 N and is at 1 = 32. Force Upper F Overscript right-arrow EndScripts Subscript 2 has a magnitude of 8 N and is at 2 = 26. In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity v Overscript right-arrow EndScripts equals left-parenthesis 14 i Overscript EndScripts minus 11 j Overscript EndScripts right-parenthesis m/s, and (c) has the v Overscript right-arrow EndScripts equals left-parenthesis 11 t i Overscript EndScripts minus 14 t j Overscript EndScripts right-parenthesis m/s2, where t is time?

Explanation / Answer

F1 = 3 (-cos 1 i + sin 1 j) N
Or F1 = 3 (- cos 32° i + sin 32° j) N
Or F1 = (-3 * cos32° i + 3 * sin 32° j) N
Or F1 = (-2.54 i + 1.59 j) N

F2 = 8 * (sin 2 i - cos 2 j) N
Or F2 = 8 * (sin 32° i - cos 32° j) N
Or F2 = (8 * sin 32° i - 8 * cos 32° j) N
Or F2 = (6.78 i - 4.24 j) N

Let third force = F3

a) If instantaneous velocity is zero i.e. velocity at just a given time is zero, then third force can be anything. But if velocity remains at zero, then velocity is constant. This means there is no acceleration. Therefore net force = 0
F1 + F2 + F3 = 0
-2.54 i + 1.59 j+ 6.78 i - 4.24 j + F3 = 0
Or 4.24 i - 2.65 j + F3 = 0
Or F3 = (-4.24 i + 2.65 j) N

b) Velocity is constant. Therefore F3 should be such that net force = 0.
Therefore, as calculated in (a),
F3 = (-4.24 i + 2.65 j) N
Note: It does not matter what is the constant velocity.

c) v = 14.0t i - 11.0 t j
Acceleration a = dv/dt = d/dt(14.0t i - 11.0 t j)
Or a = 14.0 i - 11.0 j
Net force = ma, where m = mass = 0.028 kg
Or net force = 0.028(14.0 i - 11.0 j)
= 0.392 i - 0.308 j

Therefore F1 + F2 + F3 = 0.392 i - 0.308 j
Or -2.54 i + 1.59 j+ 6.78 i - 4.24 j + F3 + F3 = 0.392 i - 0.308 j
Or 4.24 i - 2.65 j + F3 = 0.65 i - 0.70 j
Or F3 = 0.65 i - 0.70 j -4.24 i + 2.65 j
Or F3 = (-3.59 i + 1.95 j) N
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Tips:-
1. You may be wondering when to use cos and when to use sin. You also may be wondering when to use + for a component and when to use -.
If a vector makes some angle theta with a direction, then the component of that vector along that direction = (+ or -) magnitude of vector * cos(theta)
And the component of vector perpendicular to that direction = (+ or -) magnitude of vector * sin(theta)
(I will write about + or - below.)
x and y are at 90 deg.
F1 makes angle theta 1 with x axis. So, I use cos(theta 1) for x component and sin(theta 1) for y component.

F2 makes angle theta 2 with y axis. So, I use cos(theta 2) for y component and sin(theta 2) for x component.
So, now you know when to use sin or cos. Now, let us come to + or -.
As you can see, F1 is towards negative x axis and positive y axis. Therefore I put minus sign for x component and plus sign for y component.
F2 is towards positive x axis and negative y axis. Therefore I put plus sign for x component and minus sign for y component.
Right is +ve x.
Left is -ve x.
Up is +ve y.
Down is -ve y.

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