A block with a mass of 8 kg is held in equilibrium?

Question

A block with a mass of 8 kg is held in equilibrium on an incline of angle = 30.0° by the horizontal force, F

Find the magnitude of F

Find the normal force exerted by the incline on the bloc

Answer 1

Draw a picture to help you.

F cos 30 = mg sin 30

F = (8*9.8)(sin 30)/(cos 30)

Answer 2

The component of the weight of the block along the incline is equal to the component of F along the incline.

Weight of block=8g
Component of weight along the incline=8gsin30

Component of F along the incline=Fcos30

Therefore,

8gsin30=Fcos30
F=8gsin30/cos30
=39.2/.866
=45.3N

Normal force exerted by the incline on the block is equal in magnitude but opposite in the direction to the sum of the component of the weight normal to the incline and the component of F normal to the incline. This can be expressed as:

Normal force exerted by the incline of block
=8gcos30+Fsin30
=67.9+22.6
=90.5N

Answer 3

choose i ought to upload a %to assist. i visit purpose and clarify the %. so which you draw your attitude, like a ramp. placed a mass on the suitable of the ramp. Draw a line down signifying the Fg. The Fn is perpendicular to the exterior, so it relatively is going to likely be pointing N30°W. Fg gets broken down into factors, Fg parallel (Fg?) and Fg perpendicular (Fg~). no longer relatively image for perpendicular, couldnt form it so w/e. Fg? is on the backside of the triangle. Fg~ is the left and good fringe of rectangle. the attitude 30° is the attitude btw Fg and Fg~. Now locate Fg. Fg=mg Fg= -137.34 N you at the instant have your hypotenuse. From right here you will locate Fg~ and Fg||. Fg||= -sixty 8.sixty seven N Fg~= -118.ninety 8 N 2) Fg~= - Fn Fn= 118.ninety 8

Answer 4

1. Do a good picture of the problem

2. Descompone the forces, very carefully

3. Use equilibrium conditions

In X: 0=mgsen(tehta) - Fcos(theta)

F = mg tangente (theta).

In Y: N - mgcos(thta) - Fsenn(theta)
then

N = mgcos(thta) +Fsenn(theta)

Answer 5

F=W*sin(theta)/cos(theta),
and
N=W*cos(theta)+F*sin(theta),
where W is the weight of the mass, N is the normal force exerted by the incline, and theta is the angle of incline.

Answer 6

N=8gCos30+FSin30
F = 8g = 80 N

N = 40(root3+1)