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?
3 answers
Can you explain the angle? Measured from what reference?
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
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 = ?
1 answer