Physics · OCR H556 · Topic 3.2

Forces in action

Dynamics, motion with non-uniform acceleration, equilibrium, density and pressure.

  • 4 sub-topics
  • Taught in Year 12
  • 43 practice questions
  • 4 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.2.1DynamicsY12
  2. 3.2.2Motion with non-uniform accelerationY12
  3. 3.2.3EquilibriumY12
  4. 3.2.4Density and pressureY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 43 questions in this topic, free, no account needed.

Materials

Easy

Density · Year 12 · 3 marks

A metal block measures 2 cm × 3 cm × 2 cm and has mass 9.48 g.
  1. Calculate the density of the metal in kg m⁻³.
    (3)

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  1. 1 cm³ = 10⁻⁶ m³.VV=12 cm3{}= 12\text{ cm}^3=1.20×10−5 m3{}= 1.20 \times 10^{-5}\text{ m}^3
  2. ρ=mV\rho = \frac{m}{V}
  3. ρ=790 kg m−3\rho = 790\ \mathrm{kg\,m^{-3}}

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Mechanics

Medium

Moment of a force · Year 12 · 5 marks

A force of 80 N is applied to the end of a spanner of length 0.3 m. The force makes an angle of 70° with the spanner.
  1. (a)
    Define the moment of a force about a point.
    (1)
  2. (b)
    Calculate the moment of the force about the nut.
    (2)
  3. (c)
    State the angle that would give the greatest moment, and that maximum moment.
    (2)

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Part (a)1 mark

  1. Force × perpendicular distance from the point to the line of action of the force.

Part (b)2 marks

  1. 80×0.3sin⁡70∘80 \times 0.3\sin 70^\circ
  2. M=22.6 N mM = 22.6\ \mathrm{N\,m}

Part (c)2 marks

  1. 90° to the spanner.
  2. Mmax⁡=24.0 N mM_{\max} = 24.0\ \mathrm{N\,m}

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Mechanics and materials

Hard

Moments and equilibrium · Year 12 · 10 marks

Figure 1 shows a uniform horizontal beam AB of length 4.0 m and weight 40 N resting on supports at A and B. A load of 20 N hangs 1.20 m from A.
AB40 N20 N4.0 m
Figure 1
  1. (a)
    State the principle of moments.
    (2)
  2. (b)
    By taking moments about A, calculate the force from the support at B.
    (3)
  3. (c)
    Calculate the force from the support at A.
    (2)
  4. (d)
    The load is moved slowly towards B. Describe and explain how the two support forces change.
    (3)

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Part (a)2 marks

  1. The condition.For a body in equilibrium, the sum of the clockwise moments about any point
  2. Both sides of the statement are needed.equals the sum of the anticlockwise moments about that point.

Part (b)3 marks

  1. The weight of a uniform beam acts at its centre. Taking moments about A removes the force at A.RB×4.0R_B \times 4.0=40×2.00{}= 40 \times 2.00+20×1.20{}+ 20 \times 1.20
  2. Rearrange.RB=1044.0R_B = \frac{104}{4.0}
  3. RB=26.0 NR_B = 26.0\ \mathrm{N}

Part (c)2 marks

  1. The vertical forces balance.RA+RB=40+20R_A + R_B = 40 + 20
  2. RA=34.0 NR_A = 34.0\ \mathrm{N}

Part (d)3 marks

  1. Describe both.The force at B increases and the force at A decreases.
  2. Explain using moments.The moment of the load about A increases as its distance from A increases, so B must supply a larger moment.
  3. The vertical forces still balance.The total of the two support forces stays the same, equal to the total weight.

Total10

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