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# Find the acceleration?

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.

### 3 Answers by Expert Tutors

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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:

∑ Fx = 40N cos(30) = ma,   a=40N cos(30)/10 kg = 3.5 m/s²
∑ Fy = 40N sin(30) +Fn -mg=0,  Fn=mg - 40N sin(30) = 10(9.8) N - 20 N= 78 N
4.9 4.9 (230 lesson ratings) (230)
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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 :)
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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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]