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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 $\alpha$ to the horizontal, where $\t... 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 diagram, the following forces act on the package:

  1. Weight (W): Acts vertically downwards, given by W=mg=1.1imes9.8=10.78W = mg = 1.1 imes 9.8 = 10.78 N.
  2. Normal Reaction (R): Acts perpendicular to the inclined plane.
  3. Applied Force (P): Acts parallel to the plane, directed upwards along the slope.
  4. Frictional Force (F): Acts parallel to the plane, directed downwards along the slope due to the tendency of the package to slip.

Step 2

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

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Answer

Using the equilibrium condition in the vertical direction, we have:

R=Wcosα+FsinαR = W \cos \alpha + F \sin \alpha

Since F=μRF = \mu R, replacing FF gives: R=1.19.8cosα+0.5RsinαR = 1.1 \cdot 9.8 \cdot \cos \alpha + 0.5R \sin \alpha

Rearranging gives: R(10.5sinα)=10.78cosαR(1 - 0.5 \sin \alpha) = 10.78 \cos \alpha

Substituting α\alpha to find RR:

  • With tanα=34\tan \alpha = \frac{3}{4}, calculate: cosα=45,sinα=35\cos \alpha = \frac{4}{5}, \sin \alpha = \frac{3}{5}

Thus, R=10.784510.535=9.8NR = \frac{10.78 \cdot \frac{4}{5}}{1 - 0.5 \cdot \frac{3}{5}} = 9.8 N

Step 3

Find the value of P.

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Answer

Using the horizontal equilibrium condition:

P+Fcosα=RsinαP + F \cos \alpha = R \sin \alpha

Substituting F=μRF = \mu R gives: P+0.5Rcosα=RsinαP + 0.5R \cos \alpha = R \sin \alpha

Solving for PP: P=R(sinα0.5cosα)P = R(\sin \alpha - 0.5 \cos \alpha)

Substituting R=9.8R = 9.8 N, sinα=35\sin \alpha = \frac{3}{5}, and cosα=45\cos \alpha = \frac{4}{5} yields: P=9.8(350.545)=1.96NP = 9.8(\frac{3}{5} - 0.5 \cdot \frac{4}{5}) = 1.96 N

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