force acts at an angle of 30º with the horizontal,
( Up or down ?
A force of 60N acts on a box with a mass of 35.0 kg sitting on the floor. The force acts at an angle of 30º with the horizontal, and the coefficient of kinetic friction between the surfaces is 0.15.
a) What is the normal force on the box exerted by the floor?
b) What is the magnitude of the resultant acceleration of the box?
3 answers
I’m not sure that’s all that is given
well, if pushing with component down
F down on floor = 60 sin 30 + 35 * 9.81 = 30 + 343 = 373 Newtons
( Use -30 if pulling up with rope)
friction force = 0.15 *373 = 55.95
60 cos 30 = 52 N
well, no acceleration, better pull instead of push
now F down = 313 N
friction force = 0.15* 313 = 47 N
52 -47 = 5 Newtons
F = m a
5= 35 a
a = 1/7 m/s^2 = 0.143
F down on floor = 60 sin 30 + 35 * 9.81 = 30 + 343 = 373 Newtons
( Use -30 if pulling up with rope)
friction force = 0.15 *373 = 55.95
60 cos 30 = 52 N
well, no acceleration, better pull instead of push
now F down = 313 N
friction force = 0.15* 313 = 47 N
52 -47 = 5 Newtons
F = m a
5= 35 a
a = 1/7 m/s^2 = 0.143