Physics question need help immediately!!!!!!!!!!!!!?

Question

A block with a mass of 12 kg is held in equilibrium on an incline of angle 30 degrees by the horizontal force, F.

(a) Find the magnitude of F.

(b) Find the normal force exerted by the incline on the block. (disregard friction.)

Answer 1

Fx = the desired force in the horizontal (theta = 0 degree). w = W sin(30); the weight of the block acting parallel to the ramp. That's offset by f = Fx cos(30); the force pushing up along the ramp. So we can set Fx cos(30) = mg sin(30) and Fx = mg tan(30); where m = 12 kg and g = 9.81 m/sec^2. You can do the math for (a).

W = mg; the weight of the block and N = W cos(30); the normal pressing down on the block due to weight. n = Fx sin(30); the normal pressing on the block due to the horizontal force. Total normal force (NN) exerted by the ramp on the block (so there is no acceleration perpendicular to the ramp) = N + n = W cos(30) + Fx sin(30) = mg cos(30) + mg [sin(30)/cos(30)]sin(30) = mg[cos(30) + sin(30)^2/cos(30)] = mg[cos(30)^2 + sin(30^2]/cos(30) = mg/cos(30) = NN.

Note, when there is no incline (theta = 0), total normal force NN = mg = W, the weight of the block; which is a pleasing result because Fx does not act in the vertical direction at all. Also note that sin(theta)^2 + cos(theta)^2 = 1.00 always, its an identity. Finally, notice that Fx per se drops out altogether. This follows because Fx can be written in terms of weight Fx = mg tan(theta); so weight is the determining factor for normal forces along with the incline angle.

Answer 2

the force pulling the block down due to gravity is
sin(30)*12*g
so a force of equal magnitude is required to hold it.

The normal force is
cos(30)*12*g

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