A particle P of mass 6 kg lies on the surface of a smooth plane. The plane is inclined at an angle of 30° to the horizontal. The particle is held in equilibrium by a force of magnitude 49 N, acting at an angle θ to the plane, as shown in Figure 1. The force acts in a vertical plane through a line of greatest slope of the plane. (a) Show that cos θ = rac{3}{5}. (b) Find the normal reaction between P and the plane. (c) The direction of the force of magnitude 49 N is now changed. It is now applied horizontally to P so that P moves up the plane. The force again acts in a vertical plane through a line of greatest slope of the plane. Find the initial acceleration of P.

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A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

Question 4

A particle P of mass 6 kg lies on the surface of a smooth plane. The plane is inclined at an angle of 30° to the horizontal. The particle is held in equilibrium by a... show full transcript

Worked Solution & Example Answer:A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

Step 1

Show that cos θ = rac{3}{5}

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Answer

To show that cos θ = rac{3}{5}, we begin by considering the equilibrium conditions for the particle P on the inclined plane. The gravitational force acting on the particle can be resolved into two components: one parallel to the plane and one perpendicular to the plane.

The equation of equilibrium in the direction perpendicular to the plane is given by:

$49 ext{ cos } heta = 6g ext{ sin } 30°$

Here, we know that $g$ (acceleration due to gravity) is approximately $9.81 ext{ m/s}^2$ , and $ext{sin } 30° = rac{1}{2}$ . Substituting these values in:

$49 ext{ cos } heta = 6 imes 9.81 imes rac{1}{2}$

From which we determine:

$49 ext{ cos } heta = 29.43 ag{1}$

Now, considering the known value of $ext{g}$ , we can manipulate this equation to show that:

$ext{ cos } heta = rac{3}{5}$

Step 2

Find the normal reaction between P and the plane

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Answer

To find the normal reaction R between P and the plane, we need to resolve the components perpendicular to the inclined plane again. The relationship is defined as:

$R = 6g ext{ cos } 30° + 49 ext{ sin } heta$

Substituting in the known values:

$R = 6 imes 9.81 imes ext{cos} 30° + 49 ext{ sin } heta$

Evaluating this with $ext{cos } 30° ext{ and } ext{sin } heta$ gives:

$R = 90 ext{ N}$

Step 3

Find the initial acceleration of P

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Answer

When the direction of the force changes, the setup alters. The force is now applied horizontally with components acting on the particle P. The equation for the component parallel to the plane becomes:

$49 ext{ cos } 30° - 6g ext{ sin } 30° = 6a$

We can rearrange this to find the acceleration a:

$49 imes ext{cos } 30° - 6 imes 9.81 imes rac{1}{2} = 6a$

Solving for a leads to:

$a = 2.17 ext{ m/s}^2$

Thus, the initial acceleration of P is approximately 2.2 m/s².

A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

Worked Solution & Example Answer:A particle P of mass 6 kg lies on the surface of a smooth plane - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

Show that cos θ = rac{3}{5}

Find the normal reaction between P and the plane

Find the initial acceleration of P

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