An object is kept in equilibrium on an incline by means of supporting force, which is at an angle of 30 degrees o the inclined. The weight of the obbject is 500N. The incline forms an angle of 45 degrees with the horizontal. The coefficent of the friction between the surfaces in contact is 0,4. Calcualate the magnitude of the supporting force.
It is not clear if 30 degrees is above or below the incline. The following solution assumes the supporting force is above the incline. You can do a similar calculation if it is below.
First, draw a free-body diagram.
Resolve all forces along (+ downwards) the incline, and normal or perpendicular (+ into) to the incline.
Assume supporting force is F newtons at 30° above the incline.
Normal force:
N = 500cos45-Fsin30
Assume frictional force opposes downward motion, it will be negative along the incline.
Resolve all forces along the incline, the sum of which should be zero for equilibrium.
500sin45 - Fcos30 - μN=0
Substitute N=500cos45-Fsin30 and solve for F.
500sin45 - Fcos30 - μ(500cos45-Fsin30)=0
You should get F=318N approx.
If the supporting force is increased to the point of moving the object UP the incline, then the frictional force is downwards along the incline, which gives another equation:
500sin45 - Fcos30 + μ(500cos45-Fsin30)=0
you should get F=464N approx.
Any force between these two limits will maintain the objection in a stationary position.