Answers:
B.) [tex]F cos\theta=F_{n}[/tex]
C.) [tex]F sin\theta=F_{g} \pm F_{f}[/tex]
Explanation:
The image attached shows the way the force [tex]F[/tex] is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:
Net Force in X:
[tex]-F_{n}+F cos\theta=0[/tex] (1)
Where:
[tex]F_{n}[/tex] is the Normal force
[tex]F[/tex] is the magnitude of the force exerted on the block
[tex]\theta[/tex] is the angle
Net Force in Y:
[tex]F sin\theta \pm F_{f}-F_{g}=0[/tex] (2)
Where:
[tex]F_{f}[/tex] is the Friction force (it is expresed with the [tex]\pm[/tex] sign because this force may be up or down, we cannot know because the block is at rest)
[tex]F_{g}[/tex] is the gravity force
Rewrittin (1):
[tex]F cos\theta=F_{n}[/tex] (3) This is according to option B
Rewritting (2):
[tex]F sin\theta=F_{g}\pm F_{f}[/tex] (3) This is according to option C
