Resolving Forces Into Components

Resolving Forces Into Components

We have looked at resolving forces into components. There is one situation we will consider where this is particularly useful, problems involving an inclined plane. It is important because the normal force depends on the component of the gravitational force that is perpendicular to the slope.

Let us consider a block on an inclined plane. The plane is inclined at an angle \(\theta\) to the horizontal. It feels a gravitational force, \(\vec{F}_g\), directly downwards. This force can be broken into components that are perpendicular to the plane and parallel to it. This is shown here:

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We can use any coordinate system to describe the situation and the simplest thing to do is to make the \(x\)-axis of the Cartesian coordinate system align with the inclined plane. Here is the same physical situation with the coordinate system drawn in:

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This means that the components of gravitational force are aligned (parallel) with one of the axes of the coordinate system. We have shown in the figures that the angle between the horizontal and the incline is also the angle between the gravitational force and its component perpendicular to the inclined plane (this is normal to the plane). Using this angle and the fact that the components form part of a right-angled triangle we can calculate the components using trigonometry:

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Using trigonometry the components are given by:\begin{align*}{F}_{gx} & = {F}_g\sin(\theta)\\{F}_{gy} & = {F}_g\cos(\theta)\end{align*}

Example: Components of Force Due to Gravity

Question

A block on an inclined plane experiences a force due to gravity, \(\vec{F}_g\) of \(\text{137}\) \(\text{N}\) straight down. If the slope is inclined at \(\text{37}\)\(\text{°}\) to the horizontal, what is the component of the force due to gravity perpendicular and parallel to the slope?

Step 1: Components

We know that for a block on a slope we can resolve the force due to gravity, \(\vec{F}_g\) into components parallel and perpendicular to the slope. \begin{align*}{F}_{gx} & = {F}_g\sin(\theta)\\{F}_{gy} & = {F}_g\cos(\theta)\end{align*}

Step 2: Calculations

This problem is straightforward as we know that the slope is inclined at an angle of \(\text{37}\)\(\text{°}\). This is the same angle we need to use to calculate the components, therefore:\begin{align*}{F}_{gx} & = {F}_g\sin(\theta)\\ & = (\text{137})\sin(\text{37}\text{°})\\ & = \text{82,45}\text{ N}\end{align*}\begin{align*}{F}_{gy} & = {F}_g\cos(\theta)\\ & = (\text{137})\cos(\text{37}\text{°})\\ & = \text{109,41}\text{ N}\end{align*}

Step 3: Final answer

The component of \(\vec{F}_g\) that is perpendicular to the slope is \(\vec{F}_{gy}\)=\(\text{109.41}\) \(\text{N}\) in the negative \(y\)-direction.

The component of \(\vec{F}_g\) that is parallel to the slope is \(\vec{F}_{gx}\)=\(\text{82.45}\) \(\text{N}\) in the negative \(x\)-direction.

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