Question

The free-body diagram below represents a 12.0 kg wagon being pulled across a yard with a 2 N force that forms a 40.0° angle with the horizontal.

calculate the magnitude of the wagons horizontal acceleration.​

Answer 1

→ The magnitude of the wagons horizontal acceleration is 8.5755 m/s².

First, apply the Second Law of Newton in the 'y' direction. Note: The P force has a component in the axis, it is Psin(40)! Therefore:

$\Large \text { N + Psin(40) = W }\\\\\Large \text { N + 2N \times sin(40) = m \times g }\\\\\Large \text { N(1+ \times 2sin(40)) = m \times g }\\\\\Large \text { \sf N = \dfrac {12 \times 9.8}{1+2sin(40)} = 51.453 N}$

So, in the X direction:

$\Large \text { m \times a = 2N \times cos(40) }\\\\\Large \text { \sf a = \dfrac {2 \times 51.453}{12} = 8.5755 m/s^2 }$

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