A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N. The plane is inclined to the horizontal at an angle $\alpha$ where $\tan\alpha = \frac{3}{4}$ as shown in Figure 1. The line of action of the force lies in the vertical plane containing P and a line of greatest slope of the plane. The coefficient of friction between P and the plane is $\mu$. Given that P is on the point of sliding down the plane, find the value of $\mu$.

A-Level Edexcel Maths Mechanics Forces: A particle P of mass 5kg is held

A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N - Edexcel - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

Question 4

A-particle-P-of-mass-5kg-is-held-at-rest-in-equilibrium-on-a-rough-inclined-plane-by-a-horizontal-force-of-magnitude-10N-Edexcel-A-Level Maths Mechanics-Question 4-2017-Paper 1.png

A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N. The plane is inclined to the horizontal at ... show full transcript

Worked Solution & Example Answer:A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N - Edexcel - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

Step 1

Find the Force Equation: $F = \mu R$

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Answer

To determine the coefficient of friction, we start with the equation of forces acting along the incline, which is given by $F = \mu R$ . Here, $F$ is the total effective force acting parallel to the incline, and $R$ is the normal reaction.

Step 2

Resolve Forces Perpendicular to the Plane

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Answer

The normal force $R$ can be calculated from the forces acting perpendicular to the plane: $R = 10 \sin(\alpha) + 5 \cos(\alpha) \, (45.2)$

Step 3

Resolve Forces Parallel to the Plane

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Answer

Next, resolve the forces acting parallel to the incline: $F = 5 \sin(\alpha) - 2 \cos(\alpha) \, (21.4)$

Step 4

Substitute in Friction Equation

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Answer

Now substituting $R$ and $F$ into the friction equation: $\mu = \frac{5 \sin(\alpha) - 2 \cos(\alpha)}{2 \sin(\alpha) + 5 \cos(\alpha)}$

Step 5

Calculation of $\mu$

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Answer

Using the value of $\alpha$ where $\tan\alpha = \frac{3}{4}$ , we find: $\mu \approx 0.47 \, or \, 0.473$

A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N - Edexcel - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

Worked Solution & Example Answer:A particle P of mass 5kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10N - Edexcel - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

Find the Force Equation: $F = \mu R$

Resolve Forces Perpendicular to the Plane

Resolve Forces Parallel to the Plane

Substitute in Friction Equation

Calculation of $\mu$

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