Q.The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along the table, which is proportional to the resultant normal force given by the table. Forces $F_1$ and $F_2$ act at the same angle $\theta$ with the horizontal and both are just initiating the sliding motion of the block along the table. Force $F_1$ is a pulling force while the force $F_2$ is a pushing force. $F_2 > F_1$, because (A) Component of $F_2$ adds up to weight to increase the normal reaction. (B) Component of $F_1$ adds up to weight to increase the normal reaction. (C) Component of $F_2$ adds up to the opposing force. (D) Component of $F_1$ adds up to the opposing force.

[!TLDR] A pushing force's downward component increases the normal reaction, which increases the maximum friction that must be overcome, so a larger pushing force is needed. [!AN …

$F_1$ is a PULLING force applied at angle theta above horizontal: its vertical component points UPWARD, partially lifting the block and thereby REDUCING the normal reaction (and hence reducing the maximum friction $\mu N$ that must be overcome). $F_2$ is a PUSHING force applied at the same angle theta but directed into the surface: its vertical component points DOWNWARD, adding to the block's weight and INCREASING the normal reaction N, which in turn increases the maximum opposing (frictional) force the horizontal component of $F_2$ must overcome to initiate sliding. Because pushing increase …

Resolve each applied force into vertical and horizontal components and see how the vertical component alters the normal reaction (and hen …

Forgetting that friction is proportional to the NORMAL reaction (not just to weight), so any force that changes N also changes the maximum avai …