Question

1. Two forces are applied to a box on a frictionless surface. One force is 40 N [S]. The second
force is 70 N IN30 °W). What is the overall net force? (Hint: you can figure out the answer without
any calculations.)
a) 40.6 N (S59.5 °E]
b) 110 N
c) 40.6 N [N59.5 °W]
d) 40.6 N

Answer 1

The overall net force is 40.6 N [N59.5 °W]

Explanation:

Net force in x direction

F_x = 70 cos 30° - 40 = 20.62 N

Net force in y direction

F_y = 70 sin 30° = 35N

The resultant of two forces is

R = $\sqrt{F _x^2 +F _y^2}$

=  $\sqrt{20.62^2 + 35^2}$

=40.6 N

The direction of resultant is

θ= tan ^-1 ($\frac{35}{20.62}$

θ=59.5 ° (N59.5 W)