Question

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.

Answer 1

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.