$\begin{array}{1 1}45N,9.8m/s^2\\85N,9.8m/s^2\\45N,4.9m/s^2\\85N,4.9m/s^2\end{array} $

The normal force must be equal to the y-component of the gravitational force (weight). Calculating the normal force is then just a matter of finding the y-component of the gravitational force = $mg \cos 30^{\large\circ} = 85$ N

We know that the force pulling the box in the positive x direction has a magnitude of $mg \sin 30^{\large\circ}$ – which is the x-component of the weight. Using Newton’s Second Law, F = ma, we just need to solve for a which is = $g \sin\theta = 9.8 \times \sin 30^{\large\circ} = 4.9 m/s^2$

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