A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3. The plane is inclined at 20° to the horizontal. The coefficient of friction between P and the plane is 0.4. The particle is in limiting equilibrium and is on the point of moving up the plane. Calculate (a) the normal reaction of the plane on P, (b) the value of X. The force of magnitude X newtons is now removed. (c) Show that P remains in equilibrium on the plane.

A-Level Edexcel Maths Mechanics Newtons Second Law: A particle P of mass 2.5 kg rest

A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3 - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Question 4

A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane,... show full transcript

Worked Solution & Example Answer:A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3 - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Step 1

(a) the normal reaction of the plane on P

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Answer

To calculate the normal reaction (R) on the particle P, we can analyze the forces acting on it.

The weight of P can be resolved into two components:

Perpendicular to the plane: $W_n = mg imes ext{cos}(20°)$
Parallel to the plane: $W_p = mg imes ext{sin}(20°)$

The equation for the normal reaction is given by:

$R = mg imes ext{cos}(20°)$ Substituting the values, we get: $R = 2.5 imes 9.81 imes ext{cos}(20°) \ R \approx 23.0 ext{ N}$

Step 2

(b) the value of X

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Answer

Using the frictional force equation, the force of friction (F_f) can be calculated as:

$F_f = \mu R$ where ( \mu = 0.4 )

Substituting the values: $F_f = 0.4 imes (23 ext{ N}) = 9.2 ext{ N}$

In limiting equilibrium, the applied force X balances the frictional force:

$X = F_f + mg \times ext{sin}(20°)$ Substituting for mg: $X = 9.2 + 2.5 imes 9.81 imes ext{sin}(20°) \approx 18 ext{ N}$

Step 3

(c) Show that P remains in equilibrium on the plane

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Answer

After the force X is removed, we check if the conditions for equilibrium are still satisfied. The total weight acting down the slope can be calculated as:

$F = mg imes ext{sin}(20°) \approx 8.4 ext{ N}$

The maximum static friction is: $F_s = \mu R = 0.4 imes (R) \approx 9.2 ext{ N}$

Since the applied forces are:

Downward force: 8.4 N
Maximum frictional force: 9.2 N

This shows that: $F < \mu R \text{ as } 8.4 < 9.2$ Thus, P remains in equilibrium on the plane.

A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3 - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Worked Solution & Example Answer:A particle P of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude X newtons acting up a line of greatest slope of the plane, as shown in Figure 3 - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

(a) the normal reaction of the plane on P

(b) the value of X

(c) Show that P remains in equilibrium on the plane

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