In the given below figure, mass 1 does not slide with respect to the surface when the horizontal force shown is applied. Determine the magnitude of the horizontal force in both figure (a) and (b), assuming that there will not be any friction. Formula :- (F = ma for determining acceleration and add all force components (Fa / m¡+ m2+ m3 = FT / m1 ; Fa =m2g / m1 (m¡+mz+m3)) F= ma; a = Fa / m1 + m2; horizontal acceleration Fa = g tan O(m¡ + m2) %3D m, = 1.2 kg %3D m, = 1.8 kg pulley 0 = 25° m3 3.0 kg m, = 1.2 kg m, = 2.8 kg %3D (a) (b)

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Chapter1: Units, Trigonometry. And Vectors
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Question on Physics - Unit - Circular Motion

In the given below figure, mass 1 does not slide with respect to the surface when the
horizontal force shown is applied. Determine the magnitude of the horizontal force in
both figure (a) and (b), assuming that there will not be any friction.
Formula :- ( F
ma for determining acceleration and add all force components
= FT / mı ; Fa =
(Fa / m¡+ m2+ m3
m2g / m (mị+m2+m3))
F= ma; a = Fa / m¡ + m2; horizontal acceleration Fa = g tan O(m¡ + m2)
m, = 1.8 kg
m, = 1.2 kg
pulley
Fa
0 = 25°
m3 =
3.0 kg m, = 1.2 kg
m, = 2.8 kg
%3D
(a)
(b)
Transcribed Image Text:In the given below figure, mass 1 does not slide with respect to the surface when the horizontal force shown is applied. Determine the magnitude of the horizontal force in both figure (a) and (b), assuming that there will not be any friction. Formula :- ( F ma for determining acceleration and add all force components = FT / mı ; Fa = (Fa / m¡+ m2+ m3 m2g / m (mị+m2+m3)) F= ma; a = Fa / m¡ + m2; horizontal acceleration Fa = g tan O(m¡ + m2) m, = 1.8 kg m, = 1.2 kg pulley Fa 0 = 25° m3 = 3.0 kg m, = 1.2 kg m, = 2.8 kg %3D (a) (b)
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