Only two forces act on an object (mass = 3.80 kg), as in the drawing. (F = 51.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object.
F x = F · cos 45° = 51 N · 0.7071 = 36.1 N F y = F · sin 45° = 51 N · 0.7071 = 36.1 N Q = m g = 3.8 kg · 9.81 m/s² = 29.43 N R y = 36.1 N - 29.43 N = 6.67 N R x = 40 N + 36.1 N = 76.1 N R = √ (R x² + R y²) = √ ( 44.4889 + 5791.21 ) = 76.39 N F = m a a = F/ m a = 76.39 N: 3.8 kg = 20.1 m/s² The magnitude of acceleration is 20.1 m/s². α = tan ^(-1) ( R y / R x ) = 6.67 / 7.61 = 0.0876478 α = 5° Direction is α =5° counterclockwise from the + x-axis.
