What is the torque and force applied??

Question

Question 2 text 2. A rod, 1.60 m long, is pinned so it can rotate about an axis 0.500 m from the left end (thin, negligible mass rigid rod). This rod is on a level table top. Force one, a push applied to the left end of the rod, is perpendicular to the length of the rod, and has a magnitude of 20.0 N. Force 2, a push applied to the right end of the rod, is applied at an angle of 30.0 degrees outward from a perpendicular line at the end of the rod where the force is being applied, and has a magnitude of 35.0 N. Find the net torque (magnitude and direction) applied to the rod.
Question 2 answers
9.42 N*m, clockwise
29.6 N*m, Clockwise
17.8 N*m, Counterclockwise
23.3 N*m, counterclockwise

Answer 1

moment1 (torque) =0.500 x 20.0 = 10.0Nm
moment2 = F x (1.60-0.50) = 1.1F
Here, F means the vertical component of the 35.0N force. [When you draw a rectangle whose diagonally opposite corners touch a line which denotes the force vector, you will see that it has a vertical component (length) and a horizontal component (breadth)] This is because moment is derived from F x d, where F is perpendicular to d.

SO, after drawing a diagram, F = 35cos 30
With this, you can find the net torque, depending on which force is producing a torque in which direction (CW or CCW).

Hope this helps=)