# Ap Phyics

Figure shows an overhead view of a 0.024 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 6 N and is at ¥è1 = 25¢ª. Force F2 has a magnitude of 10 N and is at ¥è2 = 27¢ª. If the lemon half is stationary, what are projections of the third force on (a)x-axis, (b)y-axis? If the lemon half has constant velocity v = (13 i - 13 j) m/s, what are projections of the third force on (c)x-axis, (d)y-axis? If the lemon half has varying velocity v = (13ti - 11tj) m/s2, where t is time, what are projections of the third force on (e)x-axis, (f)y-axis?

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1. F1x = -F1•cosα = -5.44 N,
F1y = F1•sinα = 2.54 N,
F2x =F2•sinβ = 4.54 N,
F2y = - F2•cosβ = - 8.9 N.

The sum of F1 and F2 is
F12x = F1x + F2x = -5.44+4.54 = - 0.9 N,
F12y = F1y + F2y = 2.54 – 8.9 = - 6.37 N.
Since the object is at rest vector sum of all forces is zero. So
F3x = - F12x = 0.9 N,
F3y = - F12y = 6.37 N,
vector F3 = (0.9i +6.37j) N.

Similar result for the second part due to the First Newton’s Law.
For the third part in vector form
ma =F1 + F2 + F3.
a =dv/dt = 13i – 11j.
Projections are
ma(x) = F1x + F2x + F3x.
a(x) = 13á
F3x = ma(x) - F1x - F2x = 0.024•13+5.44-4.54 = 1.212 N.
ma(y) = F1y + F2y + F3y.
a(y) = - 11á
F3y = ma(y) - F1y - F2y = - 0.024•11 – 2.54+ 8.9 = 6.096 N.
vector F3 = (1.212i + 6.096j) N.

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