Question
a force F pushes on a box of mass m that initially is at rest on a rough, horizontal surface. the force F is directed at an angle theta = 30 degreees above the horizontal. the coefficants of friction between the plane and the box are static= .30 and kinetic = .28. if F=50N and m=15kg, what is the acceleration of the bos and the magnitude of the friction force?
Answers
Wb = mg = 15kg * 9.8N/kg = 147 N.
Fb = 147N @ oDeg.
Fp = 147*sin(0) = 0 = Force parallel to surface.
Fv = 147cos(0) -50*sin30 = 122 N. = Force perpendicular to surface.
Fn = Fap*cos30 - Fp - Fk,
Fn = 50*cos30 - 0 - 0.28*122,
Fn = 43.3 - 34.2 = 9.14 N. = Net force.
a = Fn/m = 9.14 / 15 = 0.609 m/s^2.
Fb = 147N @ oDeg.
Fp = 147*sin(0) = 0 = Force parallel to surface.
Fv = 147cos(0) -50*sin30 = 122 N. = Force perpendicular to surface.
Fn = Fap*cos30 - Fp - Fk,
Fn = 50*cos30 - 0 - 0.28*122,
Fn = 43.3 - 34.2 = 9.14 N. = Net force.
a = Fn/m = 9.14 / 15 = 0.609 m/s^2.