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A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

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A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle α to the horizontal, where tan α = ¾... show full transcript

Worked Solution & Example Answer:A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force - Edexcel - A-Level Maths Mechanics - Question 5 - 2009 - Paper 1

Step 1

Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.

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Answer

In the figure, the following forces are acting on the package:

  1. Weight (mg) acting vertically downwards. Here, the weight is calculated as:

    mg=1.1imes9.8extN=10.78extNmg = 1.1 imes 9.8 ext{ N} = 10.78 ext{ N}

  2. Normal force (N) acting perpendicular to the inclined plane.

  3. Frictional force (F) acting parallel to the plane, directed up the slope (opposing motion).

  4. Applied force (P) acting horizontally. The forces should be represented with appropriate arrows indicating their directions on the inclined plane.

Step 2

Find the magnitude of the normal reaction between the package and the plane.

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Answer

To find the normal reaction (N) acting on the package, we can resolve forces perpendicular to the inclined plane. The equation for equilibrium in the vertical direction is:

N = mg imes rac{1}{ ext{cos} α}

We first compute the angle α with the given tan α:

tan α = rac{3}{4}

Using trigonometric identities,

ext{cos} α = rac{4}{5}

Therefore, substituting into the equation:

N = rac{10.78 ext{ N}}{ ext{cos} α} = rac{10.78 ext{ N}}{0.8} = 13.475 ext{ N}. Thus, the magnitude of the normal reaction is approximately 13.48 N.

Step 3

Find the value of P.

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Answer

The equation for the forces acting parallel to the plane is:

P+F=RimesextsinαP + F = R imes ext{sin} α

Where:

  • F = rac{1}{2}R
  • RR is the normal reaction calculated earlier.
  • Using friction: F=extfrictioncoefficientimesN=0.5imesN=0.5imes13.475extN=6.7375extNF = ext{friction coefficient} imes N = 0.5 imes N = 0.5 imes 13.475 ext{ N} = 6.7375 ext{ N}

Now substituting this back into the equation for forces parallel to the incline:

P+6.7375=RimesextsinαP + 6.7375 = R imes ext{sin} α

The value of R from earlier is approximately equal to 13.48 N. The sine of α can also be calculated as:

ext{sin} α = rac{3}{5}

Thus:

P=RimesextsinαFP = R imes ext{sin} α - F

Finally, substituting in the numbers:

P = 13.48 imes rac{3}{5} - 6.7375 = 8.088 - 6.7375 = 1.35 ext{ N}

Therefore, the value of P is approximately 1.35 N.

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