Question
What horizontal force is needed to push a hospital bed of mass 120 kg at a constant speed up a smooth sloping plane inclined at an angle of 37 degrees above the ground?
Answers
m*g = 120kg * 9.8N/kg = 1176 N. = Wt. of
the bed.
Fp = 1176*sin37 = 707.73 N. = Force
parallel to the incline.
Fn = 1176*cos37 = 939.20 N. = Normal =
Force perpendicular to the incline.
Fap-Fp = m*a
Fap-707.73 = m*0 = 0
Fap = 707.73 N. = Force applied parallel
to the incline.
Fx = Fap*cos A = 707.73*cos37 = 565 N.
= Hor. force applied.
the bed.
Fp = 1176*sin37 = 707.73 N. = Force
parallel to the incline.
Fn = 1176*cos37 = 939.20 N. = Normal =
Force perpendicular to the incline.
Fap-Fp = m*a
Fap-707.73 = m*0 = 0
Fap = 707.73 N. = Force applied parallel
to the incline.
Fx = Fap*cos A = 707.73*cos37 = 565 N.
= Hor. force applied.
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