The figure shows an overhead wiew of a 0.021 kg lemon half and two of the three

Question

The figure shows an overhead wiew of a 0.021 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F_1 has a magnitude of 3 N and is at theta_1 - 28^+. Force F_2 has a magnitude of 6 N and is at theta_2 - 26degree. In unit vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity nu = (14t - 13j) m/s, and (c) has the nu = (11ti - 14tj) m/s^2, where t is time? (a) Number i+ j Units (b) Number i+ j Units (c) Number i+ j Units

Explanation / Answer

F1 = 3 (-cos 1 i + sin 1 j) N
Or F1 = 3 (- cos 28° i + sin 28° j) N
Or F1 = (-3 * cos28° i + 3 * sin 28° j) N
Or F1 = (-2.7 i + 1.4 j) N

F2 = 6 * (sin 2 i - cos 2 j) N
Or F2 = 6 * (sin 26° i - cos 26° j) N
Or F2 = (6 * sin 26° i - 6 * cos 26° j) N
Or F2 = (2.6 i - 5.4 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 + 1.4 + 2.6 i - 5.4 j + F3 = 0
Or-0.1 i - 4 j + F3 = 0
Or F3 = (0.1 i + 4 j) N

b) Velocity is constant. Therefore F3 should be such that net force = 0.
Therefore, as calculated in (a),
F3 = (0.1 i + 4 j) N
Note: It does not matter what is the constant velocity.

c) v = 11t i - 14.0 t j
Acceleration a = dv/dt = d/dt(11t i - 14.0 t j)
Or a = 11 i - 14.0 j
Net force = ma, where m = mass = 0.021 kg
Or net force = 0.021(11 i - 14.0 j)
= 0.23 i - 0.29 j

Therefore F1 + F2 + F3 = 0.65 i - 0.70 j
Or -2.7 i + 1.4 + 2.6 i - 5.4 j + F3 = 0.23 i - 0.29 j
Or -0.1 i - 4 j + F3 = 0.23 i - 0.29 j
Or F3 = 0.23 i - 0.29 j + 0.1 i + 4 j
Or F3 = (0.33 i + 3.71 j) N

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