Question
A traveller is pulling a 55kg load at 150degrees with a force of 600N along a horizontal floor. If the coefficient of friction between the floor and the load is 0.3, find the acceleration of the load?
Answers
bobpursley
Can you explain the angle? Measured from what reference?
Henry
Unless otherwise stated, the angle is measure CCW from the +x-axis.
150o CCW = 30o N. of W.
Mg = 55 * 9.8 = 539 N. = Wt. of load = Normal force(Fn).
Fk = u*Fn = 0.3 * 539 = 161.7 N. = Force of kinetic friction.
F*Cos30-Fk = M*a.
600*Cos30 - 161.7 = 55*a,
a = ?
F*Cos30
150o CCW = 30o N. of W.
Mg = 55 * 9.8 = 539 N. = Wt. of load = Normal force(Fn).
Fk = u*Fn = 0.3 * 539 = 161.7 N. = Force of kinetic friction.
F*Cos30-Fk = M*a.
600*Cos30 - 161.7 = 55*a,
a = ?
F*Cos30
Henry.
Correction: M*g = 55 b* 9.8 = 539 N. = Wt. of load.
Fn = 539 - 600*sin30 = 239 N.
Fk = u*Fn = 0.3 * 239 = 71.7 N. = Force of kinetic friction.
F*Cos30-Fk = M*a.
600*Cos30-71.7 = 55*a.
a = ?
Fn = 539 - 600*sin30 = 239 N.
Fk = u*Fn = 0.3 * 239 = 71.7 N. = Force of kinetic friction.
F*Cos30-Fk = M*a.
600*Cos30-71.7 = 55*a.
a = ?