_{x}= 40N cos(30) = ma, a=40N cos(30)/10 kg = 3.5 m/s²

_{y}= 40N sin(30) +F

_{n}-mg=0, F

_{n}=mg - 40N sin(30) = 10(9.8) N - 20 N= 78 N

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.

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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

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** :)

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/sec^{2})/(10 kg)]

We must cancel units where appropriate...

a=(0.866)[(40 kg-m/sec^{2})/(10 kg)]

a=(0.866)[(40 m/sec^{2})/10]

a=3.464 m/sec^{2}

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/sec^{2})/10 kg]

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