Asked by Katie
You want to use a rope to pull a 12-kg box of books up a plane inclined 30∘ above the horizontal. The coefficient of kinetic friction is 0.29. The rope pulls parallel to the incline.
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?
I found that the force you would need to use at constant speed would be 88.3 N
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?
I found that the force you would need to use at constant speed would be 88.3 N
Answers
Answered by
Henry
m*g = 12kg * 9.8N/kg = 117.6 N. = Wt. of
box.
Fp = 117.6*Sin30 = 58.8 N. = Force
parallel to the incline.
Fn = 117.6*Cos30 = 101.8 N. = Normal =
Force perpendicular to the incline.
Fk = u*Fn = 0.29 * 101.8 = 29.53 N. =
Force of kinetic friction.
Fex-Fp-Fk = m*a
Fex-58.8-29.53 = 12 * 0.50
Fex - 88.33 = 6
Fex = 94.3 N = Force exerted.
Note: The box is to be pulled with constant acceleration NOT constant velocity.
box.
Fp = 117.6*Sin30 = 58.8 N. = Force
parallel to the incline.
Fn = 117.6*Cos30 = 101.8 N. = Normal =
Force perpendicular to the incline.
Fk = u*Fn = 0.29 * 101.8 = 29.53 N. =
Force of kinetic friction.
Fex-Fp-Fk = m*a
Fex-58.8-29.53 = 12 * 0.50
Fex - 88.33 = 6
Fex = 94.3 N = Force exerted.
Note: The box is to be pulled with constant acceleration NOT constant velocity.
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