Forces and friction Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Forces and friction questions. See exactly how to solve problems on friction, horizontal, limiting, coefficient.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle of mass 4 kg4\,\text{kg} rests on a rough horizontal surface with coefficient of friction μ=0.3\mu=0.3. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the maximum possible frictional force.

Worked solution

  1. Recall the limiting friction formula

    Fmax⁡=\muRF_{\max}=\muR

    The maximum frictional force before slipping is μR\mu R.

  2. Find the normal reaction on a horizontal surface

    R=mg=4×9.8R=mg=4\times 9.8

    On a horizontal plane the normal reaction equals the weight.

  3. State the maximum frictional force

    Fmax⁡=11.76 NF_{\max}=11.76\,\text{N}

    This is the limiting value of friction.

Answer
Fmax⁡=11.76 NF_{\max}=11.76\,\text{N}
Question 2
2 markseasy
A particle of mass 5 kg5\,\text{kg} rests on a rough horizontal surface with coefficient of friction μ=0.25\mu=0.25. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the maximum possible frictional force.

Worked solution

  1. Recall the limiting friction formula

    Fmax⁡=\muRF_{\max}=\muR

    The maximum frictional force before slipping is μR\mu R.

  2. Find the normal reaction on a horizontal surface

    R=mg=5×9.8R=mg=5\times 9.8

    On a horizontal plane the normal reaction equals the weight.

  3. State the maximum frictional force

    Fmax⁡=12.25 NF_{\max}=12.25\,\text{N}

    This is the limiting value of friction.

Answer
Fmax⁡=12.25 NF_{\max}=12.25\,\text{N}
Question 3
2 markseasy
A particle of mass 10 kg10\,\text{kg} on a rough horizontal plane is pulled by a horizontal force of 24 N24\,\text{N} and is on the point of moving. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find μ\mu.

Worked solution

  1. State the condition for limiting equilibrium

    F=\muRF=\muR

    At the point of slipping, friction equals μR\mu R.

  2. Use R=mgR=mg and rearrange for μ\mu

    R=98 N, μ=2498R=98\,\text{N},\ \mu=\dfrac{24}{98}

    On a horizontal surface the normal reaction is the weight.

  3. State the coefficient of friction

    μ=2498=0.2449\mu=\dfrac{24}{98}=0.2449

    This is the coefficient of friction.

Answer
μ=0.2449\mu=0.2449
Question 4
2 markseasy
A particle of mass 2 kg2\,\text{kg} rests on a rough plane inclined at 30∘30^\circ to the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the normal reaction, to 3 significant figures.

Worked solution

  1. Resolve perpendicular to the plane

    R=mgcos⁡θR=mg\cos\theta

    The normal reaction balances the perpendicular component of the weight.

  2. Substitute the mass and angle

    R=2×9.8×cos⁡30∘R=2\times 9.8\times\cos 30^\circ

    Insert mm, gg and the angle to the horizontal.

  3. State the normal reaction

    R=17 NR=17\,\text{N}

    This is the required normal reaction to 3 s.f.

Answer
R=17 NR=17\,\text{N}
Question 5
2 markseasy
A particle of mass 3 kg3\,\text{kg} rests on a rough plane inclined at 40∘40^\circ to the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the normal reaction, to 3 significant figures.

Worked solution

  1. Resolve perpendicular to the plane

    R=mgcos⁡θR=mg\cos\theta

    The normal reaction balances the perpendicular component of the weight.

  2. Substitute the mass and angle

    R=3×9.8×cos⁡40∘R=3\times 9.8\times\cos 40^\circ

    Insert mm, gg and the angle to the horizontal.

  3. State the normal reaction

    R=22.5 NR=22.5\,\text{N}

    This is the required normal reaction to 3 s.f.

Answer
R=22.5 NR=22.5\,\text{N}

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