The figure shows an overhead view of a 0.225 kg blouse and two of the three hori

Question

The figure shows an overhead view of a 0.225 kg blouse and two of the three horizontal forces applied to it by mad Black Friday shoppers as it lies on a frictionless table. Force F_1 has a magnitude of 6.00 N and acts at theta_1 = 30 degrees Force F_1 has a magnitude of 7.00 N and acts at theta_2 = 30 degrees Determine the third force, in unit vector notation, acting on the blouse if it has the following velocities zero constant velocity v right words arrow = (6.5i with Hat - 2.6j with Hat)m/s varying velocity v(t) right words arrow = (6.5i with Hat - 2.6j with Hat)m/s where t is in serconds

Explanation / Answer

F1 = 6 (-cos 1 i + sin 1 j) N
F1 = 6 (- cos 30° i + sin 30° j) N
F1 = (-6 * cos30° i + 6 * sin 30° j) N
F1 = (-5.2 i + 3 j) N

F2 = 7 * (sin 2 i - cos 2 j) N
F2 = 7 * (sin 30° i - cos 30° j) N
F2 = (7 * sin 30° i - 7 * cos 30° j) N
F2 = (3.5 i - 6.06 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
-5.2 i + 3 j + 3.5 i - 6.06 j + F3 = 0
-1.7 i - 3.06 j + F3 = 0
F3 = (1.7 i + 3.06 j) N

b) Velocity is constant. Therefore F3 should be such that net force = 0.
Therefore, as calculated in (a),
F3 = (1.7 i + 3.06 j) N
Note: It does not matter what is the constant velocity.

c) v = 6.5t i - 2.6 t j
Acceleration a = dv/dt = d/dt(6.5t i - 2.6 t j)
a = 6.5 i - 2.6 j
Net force = ma, where m = mass = 0.225 kg
net force = 0.225(6.5 i - 2.6 j)
              = 1.4625 i - 0.585 j

Therefore F1 + F2 + F3 = 1.4625 i - 0.585 j
-5.2 i + 3 j + 3.5 i - 6.06 j + F3 = 1.4625 i - 0.585 j
- 1.7 i - 3.06 j + F3 = 1.4625 i - 0.585 j
F3 = 1.4625 i - 0.585 j + 1.7 i + 3.06 j
F3 = (3.1625 i + 2.475 j) N

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