A block of weight 30N lies on a horizontal table top. A force of 40N acts parallel to the surface of the table top, but does not cause the block to move. What is the magnitude of the frictional force between the block and the table?

ah...i remember these days of vectors...

anyway, i THINK (not so sure thou) that this is the answer...
well, if 30 N is pointed downward and 40 N is pointed sideways, i think you would need to use the Pythagorean theorem.
so...
Let Vr = (Resultant vector)
Let Vy = the the block's weight
Let Vx = the parallel force

(Vr)^2 = (Vy)^2 + (Vx)^2
Vr = squ-root( Vy^2 + Vx^2)
squ-root( 30^2 + 40^2 ) = 50 N

THEREFORE....
the magnitude of the force is 50 N
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and always remember...HERP THE DERP!!!

To find the magnitude of the frictional force between the block and the table, we'll need to apply Newton's second law of motion. In this case, since the block is not moving, we know that the net force acting on the block must be zero.

First, let's draw a free-body diagram for the block:

1. We have the weight of the block acting vertically downward with a magnitude of 30N.
2. The force applied parallel to the surface of the table has a magnitude of 40N.

Since the block is in equilibrium, the forces acting vertically and horizontally must balance each other. Therefore, the vertical forces cancel out: the weight of the block is balanced by the normal force exerted by the table.

Now, let's analyze the horizontal forces. The force applied parallel to the surface of the table and the frictional force are the only horizontal forces acting on the block. Since the block is not moving, the net force in the horizontal direction must be zero.

The force applied parallel to the surface of the table is 40N, and since it does not cause the block to move, the frictional force must have the same magnitude but in the opposite direction. So, the magnitude of the frictional force is also 40N.

Therefore, the magnitude of the frictional force between the block and the table is 40N.