The figure below shows an overhead iew of a lemon half and two of the three ho z

Question

The figure below shows an overhead iew of a lemon half and two of the three ho zontal forces that act on it as it s on a frictionless table Force F1 has a magnitude of 5 60 N and is at 1 is at 0 20°. The lemon half has mass 0.0300 kg. 2Force F2 has a magnitude of 7.00 N and physPa Operatio 8, (a) What is the third force if the lemon half has zero velocity? Rclations nVoctors (b) what is the third force if the lemon half has constant velocity v (13.0°-14.0^) m/s? le Greek (c) What is the third force if the lemon half has a varying vclocity (13.0ti 14.0ti) m's, where t is timc in scconds? 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

Explanation / Answer

Given,

F1 = 5.6 N ; theta1 = 26 deg

F2 = 7 N ; theta2 = 26 deg

In polar form

F1 = - F1 cos(theta1) i + F1 sin(theta1) j

F1 = - 5.6 x cos26 i + 5.6 x sin26 j = - 5.03 + 2.45 j

F2 = F2 sin(theta2) i - F2 cos(theta2)j

F2 = 7 sin26 i - 7 cos26 j = 3.07 i - 6.29 j

a)when v = 0 ; a = 0 ; Fnet = 0

F1 + F2 + F3 = 0

F3 = -(F1 + F2) = - (-5.03 i + 2.45 j + 3.07 i -6.29 j) = - (- 1.96 i - 3.84 j) = 1.96 i + 3.84 j

F3 = 1.96 i + 3.84 j

magnitude = sqrt (1.96^2 + 3.84^2) = 4.31 N

Hence, F3 = 4.31 N

b)constant velocity implied zero acceleration.

F3 = 1.96 i + 3.84 j

magnitude = sqrt (1.96^2 + 3.84^2) = 4.31

Hence, F3 = 4.31 N

c)v = 13 t i - 14 t j

we know that, a = dv/dt

a = 13 i - 14 j

ma = 0.03 (13i - 14j) = 0.39 i - 0.42 j

F1 + F2 + F3 = 0.39 i - 0.42 j

F3 = (0.39 i - 0.42 j) - (F1 + F2)

F3 = (0.39 i - 0.42 j) - (-5.03 i + 2.45 j + 3.07 i -6.29 j)

F3 = 0.39 i - 0.42j - (- 1.96 i - 3.84 j) = 2.35 i + 3.42 j

F3 = sqrt (2.35^2 + 3.42^2) = 4.15 N

Hence, F3 = 4.15 N

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