Mass on a frictionless incline

Problem

A block of $3\;\text{kg}$ slides on an incline at $30^\circ$ with no friction. Find the acceleration and the normal reaction.

Solution

Data. $m = 3\;\text{kg}$, $\theta = 30^\circ$, frictionless plane ($\mu = 0$).

Magnitude of the weight. $\abs{\vec{P}} = mg = 3 \cdot 9{,}81 = 29{,}43\;\text{N}$

Choice of axes and free-body diagram. Take the $x$-axis along the incline (positive pointing up the slope) and the $y$-axis perpendicular to the incline. Only the weight $\vec{P}$ and the normal reaction $\vec{R}$ (normal to the plane, no friction) act on the block. The weight has components $-mg\sin\theta$ along $x$ and $-mg\cos\theta$ along $y$.

Newton’s second law ($\vec{P} + \vec{R} = m\vec{a}$), projected onto the two axes: $\begin{cases} -mg\sin30^\circ + 0 = m\,a \\ -mg\cos30^\circ + R = 0 \end{cases}$

Acceleration (from the first equation, mass cancels): $a = -g\sin30^\circ = -9{,}81 \cdot 0{,}5 = -4{,}9\;\text{m/s}^2$ The negative sign indicates the block accelerates downwards (towards negative $x$), i.e. it slides down the slope.

Normal reaction (from the second equation): $R = mg\cos30^\circ = 29{,}43 \cdot 0{,}866 = 25{,}5\;\text{N}$

Result. $\ev{a = g\sin\theta = 4{,}9\;\text{m/s}^2 \quad (\text{down the slope}), \qquad R = mg\cos\theta = 25{,}5\;\text{N}}$

Links

Topics: Dynamics Concepts: Newton’s second law · Weight force · Normal force Skills: Resolution on an incline · Free-body diagram Objects: Inclined plane · Block