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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
the normal reaction of the plane on P
Answer
To determine the normal reaction R on the plane, we start from the forces acting in the vertical direction. The weight W of the particle P can be calculated as:
The component of the weight perpendicular to the plane acts downwards and can be expressed as:
Thus, the normal reaction R is given by:
Calculating this, we find:
R ≈ 2.5 × 9.81 × 0.9397 ≈ 23.0 ext{ N}
Therefore, the normal reaction R on P is approximately 23 N.
Step 2
the value of X.
Answer
Next, we need to find the value of X. The frictional force F_f can be calculated using the friction coefficient and the normal reaction:
Substituting R from part (a):
= 9.2 ext{ N}$$ In equilibrium, the force acting parallel to the inclined plane (X) must balance the weight component acting down the slope. The weight component down the slope is: $$W_{parallel} = mg imes ext{sin}(20°) = 2.5 imes 9.81 imes ext{sin}(20°)$$ Calculating this gives: $$W_{parallel} ≈ 8.38 ext{ N}$$ Thus, using the equilibrium condition: $$X = W_{parallel} + F_f = 8.38 + 9.2 = 17.58 ext{ N}$$ Hence, rounding gives X ≈ 18 N.Step 3
Show that P remains in equilibrium on the plane.
Answer
After removing the force X, we need to show that P remains in equilibrium. The only forces acting in the direction of the plane are the weight component down the slope and the friction. Now, the force down the slope is:
The maximum static friction is calculated as:
Since the downward force (8.38 N) is less than the maximum static friction (9.2 N):
This ensures that P remains in equilibrium, confirming that it will not move down the plane when X is removed.