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!!!
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?
1 answer
1 answer