The figure shows an overhead view of a 0.027 kg lemon half and two of the three

Question

The figure shows an overhead view of a 0.027 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 = 25. Force Upper F Overscript right-arrow EndScripts Subscript 2 has a magnitude of 7 N and is at 2 = 32. 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 15 j Overscript EndScripts right-parenthesis m/s, and (c) has the v Overscript right-arrow EndScripts equals left-parenthesis 14 t i Overscript EndScripts minus 14 t j Overscript EndScripts right-parenthesis m/s2, where t is time?

Explanation / Answer

For simplicity, I am writing i and j instead of ihat and jhat.

F1 = 3 (-cos 1 i + sin 1 j) N
Or F1 = 3 (- cos 25° i + sin 25° j) N
Or F1 = (-3 * cos25° i + 3 * sin 25° j) N
Or F1 = (-2.7 i + 1.3 j) N

F2 = 7 * (sin 2 i - cos 2 j) N
Or F2 = 7 * (sin 32° i - cos 32° j) N
Or F2 = (7 * sin 32° i - 7 * cos 32° j) N
Or F2 = (3.7 i – 5.9 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
(-2.7 i + 1.3 j + 3.7 i – 5.9 j + F3 = 0
Or 1.0 i – 4.6 j + F3 = 0
Or F3 = (-1.0 i + 4.6 j) N

it has magnitude = (1.0² + 4.6²) N = 4.71 N
and direction = arctan(4.6/-1.0) = 77.7º add 180º to place in Q3 = 257.7º ccw from +x

b) Velocity is constant. Therefore F3 should be such that net force = 0.
Therefore, as calculated in (a),
F3 = (-1.0 i + 4.6 j) N
Note: It does not matter what is the constant velocity.

c) v = 14.0t i - 14.0 t j
Acceleration a = dv/dt = d/dt(14.0t i - 14.0 t j)
Or a = 14.0 i - 14.0 j
Net force = ma, where m = mass = 0.027 kg
Or net force = 0.027(14.0 i - 14.0 j)
= 0.378 i - 0.378 j

Therefore F1 + F2 + F3 = 0.378 i - 0.378 j
Or -2.7 i + 1.3 j + 3.7 i – 5.9 j + F3 = 0.378 i - 0.378 j
Or 1.0 i – 4.6 j + F3 = 0.378 i - 0.378 j
Or F3 = 0.378 i - 0.378 j - 1.0 i + 4.6 j
Or F3 = (-0.622 i + 4.22 j) N

it has magnitude = (0.622 ² + 4.22²) N = 4.26 N
and direction = arctan(4.22/-0.622) = 81.61º

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