Find the acceleration?

Question

A 10 kg box is resting on a smooth (frictionless) horizontal surface of a table. You pull the box by the attached ribbon with a force of 40 N at a 30 degrees angle from the surface of the table. Find the acceleration of the box and the magnitude of the normal force. Assume that friction can be neglected.

Answer 1

Draw a free-body diagram and identify all forces.

Find the sum of all forces in the horizontal (x) and vertical (y) direction and use Newton's 2nd law:

 

∑ F_x = 40N cos(30) = ma,   a=40N cos(30)/10 kg = 3.5 m/s²

∑ F_y = 40N sin(30) +F_n -mg=0,  F_n=mg - 40N sin(30) = 10(9.8) N - 20 N= 78 N

Answer 2

Hey Sun -- 40N at 30 deg provides 20N lift and a 35N horizontal ...

the 100N box "feels" 20N lighter at 80N ... 35N tugs a 10kg load at 3.5 m/s/s :)

Answer 3

Hi Sun;

F=ma

a=F/m

The force is being applied horizontally.  A triangle is created by the ribbon, table-surface, and angle of 30-degrees.  We need to establish the cos of 30-degrees.  This will give us the value of the adjacent-side/hypotenuse.

a=(cos 30-degrees)(F/m)

a=(0.866)[(40 kg-m/sec²)/(10 kg)]

We must cancel units where appropriate...

a=(0.866)[(40 kg-m/sec²)/(10 kg)]

a=(0.866)[(40 m/sec²)/10]

a=3.464 m/sec²

 

Another approach to this is as follows...

The triangle formed has three angles: 30, 60 and 90 degrees.

The adjacent side would be proportionately, the square-root of 3.  The hypotenuse would be 2.  (The opposite would be 1.)

Therefore...

a=[(√3)/(2)][(40 kg-m/sec²)/10 kg]