Wb = 2.8kg * 9.8N/kg = 27.44 N. = Wt. of
block.
Fb = 27.44N.[56.6o] = Force of block.
Fp = 27.44Sin56.6 = 22.91 N. = Force
parallel to the incline.
a. cos56.6 = F/27.44
F = 27.44cos56.6 = 15.11 N.
b. Fv = 27.44cos56.6 = 15.11 N. = Force perpendicular to the incline = normal.
A block of mass m = 2.80kg is held in
equilibrium on a frictionless incline of angle= 56.6 by the horizontal force F. (a) Determine the value of F.(b) Determine the normal force exerted by the incline on the block. Hint: Draw a free-body
diagram of the block. Choose a coordinate system with the x axis parallel to the surface of the incline
and the y axis perpendicular to the surface of the incline.
1 answer
1 answer