Wb = m*g = 35kg * 9.8N/kg = 343 N. = Wt.
of box.
Fb = 343N @ 0o.
Fp=343*sin(0) = 0.=Force parallel to bed
Fv = 343*cos(0) = 343 N. = Force perpendicular to bed.
Fs = u*Fv = 0.3*343 = 102.9 N. = Force of static friction.
A. Fn = Fp - Fs = m*a.
0 - 102.9 = 35*a
qa = -2.94 m/s^2.
B. Fk = u*Fv = 0.1*343 = 34.3 N.
Fp - Fk = m*a.
0-34.3 = 35*a
a = -0.98 m/s^2.
A 35 kg box rest in the bed of a truck. The coefficient of static friction between the box and the truck bed is 0.30 and the coefficient of friction of kinetic friction is 0.10.
A) Calculate the maximum acceleration the truck can have before the box slides backwards.
B) The truck just exceeds the maximum acceleration that prevents the box from sliding backwards. Calculate the acceleration of the box.
2 answers
сука Блядь rush B
2 answers