a block of wood weighing 18 pounds rests on an inclined plane that makes an angle of 60 with the horizontal. how much of the friction force is needed to prevent the block from sliding on the plane?

501.46 pound force of frictional force will be needed.

Step-by-step explanation:

As shown in the figure attached, a wooden block has been placed on an inclined plane that makes an angle of 60° from the horizontal.

Block is placed on an inclined surface therefore block will tend to slide because of a component of its weight.

This component of weight will be = mgsin60°

Where m = 18 pounds

and g = 32.17 ft per sec²

Now to balance this component frictional force will be required equal in measure but in opposite direction.

Therefore measure of friction = 18×32.17×sin60°

= 18×32.17×0.866

= 501.46 pound force

Therefore, 501.46 pound force of frictional force will be needed.

1/2*qr*rs

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)