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Object on an Inclined Plane

Object on an Inclined Plane

In an earlier section we looked at the components of the gravitational force parallel and perpendicular to the slope for objects on an inclined plane. When we look at problems on an inclined plane we need to include the component of the gravitational force parallel to the slope.

Think back to the pictures of the book on a table, as one side of the table is lifted higher the book starts to slide. Why? The book starts to slide because the component of the gravitational force parallel to the surface of the table gets larger for the larger angle of inclination. This is like the applied force and it eventually becomes larger than the frictional force and the book accelerates down the table or inclined plane.

The force of gravity will also tend to push an object ‘into’ the slope. This is the component of the force perpendicular to the slope. There is no movement in this direction as this force is balanced by the slope pushing up against the object. This “pushing force” is the normal force (N) which we have already learnt about and is equal in magnitude to the perpendicular component of the gravitational force, but opposite in direction.

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Tip:

Do not use the abbreviation \(W\) for weight as it is used to abbreviate ‘work’. Rather use the force of gravity \({F}_{g}\) for weight.

Example: Newton’s Second Law: Box on Inclined Plane

Question

A body of mass \(M\) is at rest on an inclined plane due to friction.

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What of the following option is the magnitude of the frictional force acting on the body?

  1. \(F_g\)
  2. \(F_g\cos(θ)\)
  3. \(F_g\sin(θ)\)
  4. \(F_g\tan(θ)\)

Step 1: Analyse the situation

The question asks us to determine the magnitude of the frictional force. The body is said to be at rest on the plane, which means that it is not moving and therefore the acceleration is zero. We know that the frictional force will act parallel to the slope. If there were no friction the box would slide down the slope so friction must be acting up the slope. We also know that there will be a component of gravity perpendicular to the slope and parallel to the slope. The free body diagram for the forces acting on the block is:

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Step 2: Determine the magnitude of the frictional force

We can apply Newton’s second law to this problem. We know that the object is not moving so the resultant acceleration is zero. We choose up the slope to be the positive direction. Therefore: \begin{align*} \vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\ F_f – F_g\sin(\theta) &= m (0)\\ F_f – F_g\sin(\theta) &= m (0)\\ F_f & = F_g\sin(\theta) \end{align*}

Step 3: Quote your final answer

The force of friction has the same magnitude as the component of the force of gravitation parallel to the slope, \(F_g\sin(\theta)\).

Example: Newton’s Second Law: Object on an Incline

Question

A force of magnitude \(T=\text{312} \text{N}\) up an incline is required to keep a body at rest on a frictionless inclined plane which makes an angle of \(\text{35}\)\(\text{°}\) with the horizontal. Calculate the magnitudes of the force due to gravity and the normal force, giving your answers to three significant figures.

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Step 1: Find the magnitude of \(\vec{F}_g\)

We are usually asked to find the magnitude of \(\vec{T}\), but in this case \(\vec{T}\) is given and we are asked to find \(\vec{F}_g\). We can use the same equation. \(T\) is the force that balances the component of \(\vec{F}_g\) parallel to the plane (\({F}_{gx}\)) and therefore it has the same magnitude.

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We can apply Newton’s second law to this problem. We know that the object is not moving so the resultant acceleration is zero. We choose up the slope to be the positive direction. Therefore: \begin{align*} \vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\ T – F_g\sin(\theta) &= m (0)\\ F_g & = \frac{T}{\sin(\theta)} \\ & = \frac{\text{312}}{\sin(\text{35}\text{°})} \\ & = \text{543,955}\text{ N} \end{align*}

Step 2: Find the magnitude of \(\vec{N}\)

We treat the forces parallel and perpendicular to the slope separately. The block is stationary so the acceleration perpendicular to the slope is zero. Once again we can apply Newton’s second law of motion. We choose the direction of the normal force as the positive direction. \begin{align*} \vec{F}_R &= m\vec{a}\; \text{using signs for direction}\\ N – F_g\cos(\theta) &= m (0)\\ N & = F_g\cos(\theta) \end{align*}

We could substitute in the value of \(F_g\) calculated earlier. We would like to illustrate that there is another approach to adopt to ensure you get the correct answer even if you made a mistake calculating \(F_g\). \(F_g\cos(\theta)\) can also be determined with the use of trigonometric ratios. We know from the previous part of the question that \(T = F_g\sin(\theta)\). We also know that \begin{align*} \tan(\theta) &=\frac{F_g\sin(\theta)}{F_g\cos(\theta)}\\ &=\frac{T}{N}\\ N&=\frac{T}{\tan(\theta)}\\ & = \frac{\text{312}}{\tan(\text{35}\text{°})} \\ & = \text{445,58}\text{ N} \end{align*}

Note that the question asks that the answers be given to 3 significant figures. We therefore round \(\vec{N}\) from \(\text{445.58}\) \(\text{N}\) up to \(\text{446}\) \(\text{N}\) perpendicular to the surface upwards and \(\vec{T}\) from \(\text{543.955}\) \(\text{N}\) up to \(\text{544}\) \(\text{N}\) parallel to the plane, up the slope.

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