The figure below shows an overhead view of a lemon half and two of the three hor

Question

The figure below shows an overhead view of a lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F^vector_1 has a magnitude of 4.60 N and is at theta_1=25 degree. Force F^vector_2 has a magnitude of 7.00 N and is at theta_2=25 degree. The lemon half has mass 0.0150 kg. (a) What is the third force if the lemon half has zero velocity? F^vector_3= N (b) What is the third force if the lemon half has constant velocity v^vector=(13.0i^^- 14.0j^^) m/s? F^vector_3= N (c) What is the third force if the lemon half has a varying velocity v^vector=(13.0ti^^- 14.0tj^^) m/s, where t is time in seconds? F^vector_3= N Acceleration is the time derivative of the velocity. Acceleration is related to the net force by Newton's second law. The net force is the vector sum of the three applied forces. So, the sum of the three forces equals the product of mass and acceleration.

Explanation / Answer

a)

F1 = - 4.60 Cos25 i + 4.60 Sin25 j = - 4.2 i + 1.94 j

F2 = 7 Cos25 i - 7 Sin25 j = 6.34 i - 2.96 j

for zero velocity

F1 + F2 + F3 = 0

(- 4.2 i + 1.94 j ) + (6.34 i - 2.96 j ) + F3 = 0

F3 = - 2.14 i + 1.02 j

magnitude : = sqrt((-2.14)2 + (1.02)2) = 2.4 N

b)

still the net force is 0

hence

magnitude = 2.4 N

c)

a = acceletaion = dv/dt = 13 i - 14 j

F1 + F2 + F3 = ma

(- 4.2 i + 1.94 j ) + (6.34 i - 2.96 j ) + F3 = (0.015) (13 i - 14 j)

(- 4.2 i + 1.94 j ) + (6.34 i - 2.96 j ) + F3 = (0.195 i - 0.21 j)

F3 = - 1.95 i + 0.81 j

magnitude : = sqrt((- 1.95)2 + (0.81)2) = 2.11 N

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