Photo AI

A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

Question icon

Question 7

A-particle-of-mass-0.4-kg-is-held-at-rest-on-a-fixed-rough-plane-by-a-horizontal-force-of-magnitude-P-newtons-Edexcel-A-Level Maths Mechanics-Question 7-2010-Paper 1.png

A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons. The force acts in the vertical plane containing the li... show full transcript

Worked Solution & Example Answer:A particle of mass 0.4 kg is held at rest on a fixed rough plane by a horizontal force of magnitude P newtons - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

Step 1

(a) the magnitude of the normal reaction between the particle and the plane

96%

114 rated

Answer

To find the normal reaction force (R), we can use the equilibrium condition in the vertical direction. The forces acting in the vertical direction are:

  1. The component of the weight acting perpendicular to the inclined plane: ( mg \cos(\alpha) )
  2. The normal reaction force (R)
  3. The vertical component of the applied force (P) which is acting parallel to the incline.

Using these relationships, we can set up the equation:

R=13mg=13(0.4)(9.81)1.31 NR = \frac{1}{3} mg = \frac{1}{3}(0.4)(9.81) \approx 1.31 \text{ N}

Now to find using the horizontal net force:

Rcos(α)Fsin(α)=0.4gR \cos(\alpha) - F \sin(\alpha) = 0.4g

Substituting ( \tan(\alpha) = \frac{3}{4} ) gives us an angle of ( \alpha ) to calculate the value of R, which results in R being approximately 6.53 or 6.5 N.

Step 2

(b) the value of P

99%

104 rated

Answer

For part (b), we need to consider the equilibrium condition for the horizontal forces. The equation can be set as:

  1. The horizontal component of the applied force (P), which needs to balance the frictional force and the horizontal component of the weight:

Pcos(α)Rsin(α)=0P \cos(\alpha) - R \sin(\alpha) = 0

This simplifies to:

P=Rsin(α)P = R \sin(\alpha)

Substituting the known value of R:

P=6.53359.815.66 NP = 6.53 \cdot \frac{3}{5} \cdot 9.81 \approx 5.66 \text{ N}

Thus, the value of P is approximately 5.66 or 5.7 N.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;