A boy pulls a box of mass 30 kg with a force of 25N at 30 degrees above the horizontal.?

Question

1 Ignoring friction, what is the acceleration of the box? 2 What is the normal force exerted on the box by the ground?

Answer 1

break up into components -.-
a = f_x/m f_x = (25N)(cos30) so a = ((25N)(cos30))/ 30kg

the normal force should be = to the weight. so it is F_n = mg = (30kg)(-9.8m/s^2)

i dont have calculator so just do it...



u look for acceleration in the direction it is moving. Since the force is applied at an angle use only the force along the x axis
and for the normal force, it is resting on the ground so the only force to be applying on the box is the ground.

Answer 2

F = ma
a = F/m = 25 cos30/ 30 = 0.7217m/s^2
Note: The normal force is diminished by the vertical component of the force that the boy exerts:
F(normal) = 30(9.8) - 25 sin30 = 281.5N