WEBVTT
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The figure shows a body resting on a rough inclined plane where the coefficient of static friction between the body and the plane, π sub π , equals 0.487.
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Given that the body is on the point of sliding down the plane, find the angle of inclination π, rounding your answer to the nearest minute, if necessary.
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Weβre told in this statement that the coefficient of static friction is 0.487.
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And weβre asked to solve for the angle of inclination π at which this inclined plane is raised.
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We can start on our solution by considering the forces that are acting on the body.
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We know that gravity acts on this body.
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We can draw that in as a vector pointing straight down and assign that force of magnitude equal to the weight of the body.
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Weβll call it capital π.
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There is also a normal force that acts on the body perpendicular to the surface of the plane.
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We can refer to that force as πΉ sub π.
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And finally, there is a frictional force acting on this body that points up the plane.
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We can call that force πΉ sub π.
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Weβre told in the statement that this body is at rest.
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Itβs not in motion.
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That means the three forces weβve identified all balance one another out.
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Their sum is zero.
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To look more closely at those forces, letβs put two coordinate axes on this scenario.
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Weβll define motion in the positive π¦-direction as motion up perpendicular to the plane and motion in the positive π₯-direction is up the plane and parallel to it.
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Two of our forces, πΉ sub π and πΉ sub π, are already completely aligned with these axes.
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πΉ sub π, however, has components in the π₯- and π¦-direction.
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If we sketch in those π₯- and π¦-components, we see that those components, along with the magnitude of the vector itself, form a right triangle.
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And the topmost angle in that triangle is equal to π.
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Now that all three of our forces are broken up into π₯- and π¦-components, letβs consider those components along the π₯-axis.
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We have the frictional force acting in the positive π₯-direction minus the π₯-component of the weight force which weβll call π sub π₯.
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That difference is equal to zero because these are all the forces that act in the π₯-direction.
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And our body is in equilibrium.
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We can expand on both these expressions, πΉ sub π and π sub π₯.
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Letβs consider π sub π₯ first.
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Looking at our diagram, we see that this component of the weight force is equal to π times the sine of π.
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So we insert that expanded expression in our equation.
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For πΉ sub π, we can recall a definition for frictional force.
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Friction force πΉ sub π is equal to the coefficient of friction π, whether static or kinetic, times the normal force πΉ sub π.
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In our case, we can write π sub π because our body is at rest not in motion.
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We can expand our πΉ sub π term further based on our diagram.
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If we consider only forces in the π¦-direction on our diagram, we see that there are two, πΉ sub π, the normal force, and the π¦-component of the weight force.
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Since our body is at rest, that must mean that the magnitude of these two forces are equal.
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In other words, πΉ sub π is equal to π times the cosine of π.
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We now have a fully expanded equation for forces in the π₯-direction of our scenario.
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If we add π times the sign of π to both sides, then we see we can cancel out the weight force, π, of our body from this expression.
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Our result is independent of it.
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At this stage, it will be helpful to recall a trigonometric identity.
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The tangent of an angle π is equal to the sine of that same angle divided by the cosine of that angle.
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So if we divide both sides of our equation by the cosine of π, then that term cancels out on the left-hand side.
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And on the right-hand side, using our identity, we have the tangent of π.
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If we then take the inverse tangent of both sides of this equation, we see that π equals the arc tangent of our coefficient of static friction, π sub π .
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Weβre given the value of π sub π and can insert that now.
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When we evaluate this expression on our calculator, we find that, to the nearest minute, π is 25 degrees and 58 minutes.
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Thatβs the maximum angle of inclination of our plane at which the body is still stationary.