The figure shows an overhead view of a 0.025 kg lemon half and two of the three

Question

The figure shows an overhead view of a 0.025 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force Upper F Overscript right-arrow EndScripts Subscript 1 has a magnitude of 6 N and is at ?1 = 27?. Force Upper F Overscript right-arrow EndScripts Subscript 2 has a magnitude of 10 N and is at ?2 = 26?. In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity v Overscript right-arrow EndScripts equals left-parenthesis 14 i Overscript ? EndScripts minus 16 j Overscript ? EndScripts right-parenthesis m/s, and (c) has the v Overscript right-arrow EndScripts equals left-parenthesis 10 t i Overscript ? EndScripts minus 12 t j Overscript ? EndScripts right-parenthesis m/s^2, where t is time?

Explanation / Answer

a) acc is zero .

so Fnet =   0

-6cos27i + 6sin27 j + 10sin26i - 10cos26j + F3 = 0

F3 = -0.96 i - 6.26 j

magnitude = sqrt(0.96^2 + 6.26^2) = 6.33 N

angle = tan-1(0.96 /6.26) = 8.72 degrees below - ve x-axis.

b) speed constant. so acc. is zer0 .

Fnet will be same as calculated in 1st part.

F3 = -0.96 i - 6.26 j

c) a = dv/dt   = 10 i + 12 j

Fnet = ma

-6cos27i + 6sin27 j + 10sin26i - 10cos26j + F3 = 0.025(10i - 12j)

F3 = 1.21 i + 5.96 j

manitude = sqrt(I^2 + j^2) = 6.08 N

direction = tan-1(1.21 / 5.96) =11.48 derees from x-axis

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