Hard Further Maths Statics of rigid bodies Questions

Challenging, exam-style Further Maths Statics of rigid bodies questions with worked solutions. Stretch yourself on the hardest hinged-rod, hinge-reaction, moments, ladder problems.

hinged-rodhinge-reactionmomentsladderrough-walllimiting-equilibrium
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A uniform solid cuboid of width 88 m and height 99 m stands on a rough plane, the coefficient of friction being 0.90.9. The inclination of the plane is slowly increased from zero. Determine whether the cuboid slides or topples first.
Show worked solution

Worked solution

  1. Write the critical angle for sliding

    tan⁡λ=μ=0.9\tan\lambda=\mu=0.9

    Sliding starts when the inclination reaches arctan⁡μ\arctan\mu.

  2. Write the critical angle for toppling

    tan⁡β=89=0.8889\tan\beta=\frac{8}{9}=0.8889

    Toppling starts when the inclination reaches arctan⁡ab\arctan\frac{a}{b}.

  3. Compare the two critical tangents

    μ>89\mu>\frac{8}{9}

    The smaller critical angle is reached first.

  4. State the conditions for the equilibrium of a rigid body

    ΣFx=0,ΣFy=0,ΣM=0\Sigma F_x=0,\quad\Sigma F_y=0,\quad\Sigma M=0

    A rigid body is in equilibrium when the forces and the moments both balance.

  5. Recall the definition of the moment of a force

    M=FdM=Fd

    The moment is the force multiplied by the perpendicular distance to the pivot.

  6. Recall the law of friction at a rough contact

    F≤μRF\le\mu R

    The friction can take any value up to μR\mu R; at the point of slipping it equals μR\mu R.

  7. Note that the weight of a uniform body acts at its centre

    W acts at the midpointW\text{ acts at the midpoint}

    A uniform rod or lamina has its centre of mass at its geometric centre.

  8. Recall that a smooth contact has no friction

    F=0 at a smooth contactF=0\text{ at a smooth contact}

    A smooth surface can only push at right angles to itself.

  9. Choose a pivot that removes an unknown force

    take moments about the line of an unknown\text{take moments about the line of an unknown}

    Taking moments through an unknown makes that force contribute no moment.

  10. Resolve the forces into horizontal and vertical components

    components: Fcos⁡θ, Fsin⁡θ\text{components: }F\cos\theta,\ F\sin\theta

    Each force is split along two perpendicular directions before balancing.

  11. Recall that the perpendicular distance uses the sine of the angle

    d=ℓsin⁡θd=\ell\sin\theta

    The moment arm of a vertical force about the foot of an inclined rod is the horizontal offset.

  12. Recall the tangent of an angle from a right triangle

    tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

    A Pythagorean triple fixes the sine and cosine exactly.

  13. State the modelling assumptions

    rigid body, light string, uniform where stated\text{rigid body, light string, uniform where stated}

    These are the standard assumptions behind every statics calculation.

  14. Note that the reaction at a hinge has two components

    R=H2+V2R=\sqrt{H^{2}+V^{2}}

    A hinge (or pin) can push in any direction, so it has a horizontal and a vertical part.

  15. Recall how to combine perpendicular components into a magnitude

    ∣R∣=H2+V2|\mathbf{R}|=\sqrt{H^{2}+V^{2}}

    The resultant of two perpendicular components is found by Pythagoras.

  16. Recall how to find the direction of a resultant

    tan⁡ϕ=VH\tan\phi=\frac{V}{H}

    The angle to the horizontal is the arctangent of the vertical over the horizontal part.

  17. Check that the three equilibrium equations are consistent

    3 equations, 3 unknowns\text{3 equations, 3 unknowns}

    Coplanar equilibrium gives exactly three independent scalar equations.

  18. State the conclusion

    toppling first\text{toppling first}

    The block topples before it slides.

Answer
topples first\text{topples first}
Question 2
9 markschallenging
A uniform rod ABAB of weight 260260 N is freely hinged at AA to a vertical wall. The rod is held in a horizontal position by a light inextensible string attached to the end BB and to a point of the wall above AA, the string making an angle α\alpha with the rod, where tan⁡α=724\tan\alpha=\frac{7}{24}. A particle of weight 110110 N hangs from the end BB. Find the magnitude of the force exerted by the hinge on the rod at AA. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Take moments about the hinge AA

    Tsin⁡α⋅L=260⋅12L+110⋅LT\sin\alpha\cdot L=260\cdot\tfrac{1}{2}L+110\cdot L

    The hinge reaction has no moment about AA, so it is the natural pivot.

  2. Solve for the tension

    T=12×260+110sin⁡α=857.1 NT=\frac{\tfrac{1}{2}\times 260+110}{\sin\alpha}=857.1\ \text{N}

    The length of the rod cancels, leaving the tension in terms of sin⁡α\sin\alpha.

  3. Resolve horizontally for the hinge reaction

    H=Tcos⁡α=822.9 NH=T\cos\alpha=822.9\ \text{N}

    The horizontal hinge component balances the horizontal pull of the string.

  4. Resolve vertically for the hinge reaction

    V=260+110−Tsin⁡α=130 NV=260+110-T\sin\alpha=130\ \text{N}

    The vertical hinge component makes up the rest of the load.

  5. Combine the components by Pythagoras

    R=H2+V2=822.92+1302R=\sqrt{H^{2}+V^{2}}=\sqrt{822.9^{2}+130^{2}}

    The hinge force is the resultant of its two perpendicular components.

  6. Note the sense of the hinge components

    H towards the wall, V upwardsH\text{ towards the wall},\ V\text{ upwards}

    Both components come out positive, confirming the assumed directions.

  7. State the conditions for the equilibrium of a rigid body

    ΣFx=0,ΣFy=0,ΣM=0\Sigma F_x=0,\quad\Sigma F_y=0,\quad\Sigma M=0

    A rigid body is in equilibrium when the forces and the moments both balance.

  8. Recall the definition of the moment of a force

    M=FdM=Fd

    The moment is the force multiplied by the perpendicular distance to the pivot.

  9. Recall the law of friction at a rough contact

    F≤μRF\le\mu R

    The friction can take any value up to μR\mu R; at the point of slipping it equals μR\mu R.

  10. Note that the weight of a uniform body acts at its centre

    W acts at the midpointW\text{ acts at the midpoint}

    A uniform rod or lamina has its centre of mass at its geometric centre.

  11. Recall that a smooth contact has no friction

    F=0 at a smooth contactF=0\text{ at a smooth contact}

    A smooth surface can only push at right angles to itself.

  12. Choose a pivot that removes an unknown force

    take moments about the line of an unknown\text{take moments about the line of an unknown}

    Taking moments through an unknown makes that force contribute no moment.

  13. Resolve the forces into horizontal and vertical components

    components: Fcos⁡θ, Fsin⁡θ\text{components: }F\cos\theta,\ F\sin\theta

    Each force is split along two perpendicular directions before balancing.

  14. Recall that the perpendicular distance uses the sine of the angle

    d=ℓsin⁡θd=\ell\sin\theta

    The moment arm of a vertical force about the foot of an inclined rod is the horizontal offset.

  15. Recall the tangent of an angle from a right triangle

    tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

    A Pythagorean triple fixes the sine and cosine exactly.

  16. State the answer

    R=833 NR=833\ \text{N}

    This is the required quantity.

Answer
R=833 NR=833\ \text{N}
Question 3
9 markschallenging
A uniform ladder of weight 480480 N rests with its foot on rough horizontal ground and its top against a rough vertical wall. The coefficient of friction at the ground is 0.50.5 and the coefficient of friction at the wall is 0.40.4. The ladder is on the point of slipping. Find the angle that the ladder makes with the horizontal. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Resolve horizontally

    μ1N1=N2\mu_1 N_1=N_2

    The friction at the ground balances the horizontal reaction of the wall.

  2. Resolve vertically

    N1+μ2N2=480N_1+\mu_2 N_2=480

    The weight is carried by the ground reaction and the friction at the wall.

  3. Solve the two resolution equations for the reactions

    N1=4801+0.5×0.4,N2=0.5N1N_1=\frac{480}{1+0.5\times 0.4},\quad N_2=0.5 N_1

    Substituting one into the other gives both normal reactions.

  4. Take moments about the foot of the ladder

    N2sin⁡θ+μ2N2cos⁡θ=12480cos⁡θN_2\sin\theta+\mu_2 N_2\cos\theta=\tfrac{1}{2}480\cos\theta

    The wall reaction and its friction both have moments about the foot.

  5. Divide by cos⁡θ\cos\theta and substitute the reactions

    tan⁡θ=1−0.5×0.42×0.5=0.8\tan\theta=\frac{1-0.5\times 0.4}{2\times 0.5}=0.8

    All the weights cancel, leaving a formula for tan⁡θ\tan\theta.

  6. State the conditions for the equilibrium of a rigid body

    ΣFx=0,ΣFy=0,ΣM=0\Sigma F_x=0,\quad\Sigma F_y=0,\quad\Sigma M=0

    A rigid body is in equilibrium when the forces and the moments both balance.

  7. Recall the definition of the moment of a force

    M=FdM=Fd

    The moment is the force multiplied by the perpendicular distance to the pivot.

  8. Recall the law of friction at a rough contact

    F≤μRF\le\mu R

    The friction can take any value up to μR\mu R; at the point of slipping it equals μR\mu R.

  9. Note that the weight of a uniform body acts at its centre

    W acts at the midpointW\text{ acts at the midpoint}

    A uniform rod or lamina has its centre of mass at its geometric centre.

  10. Recall that a smooth contact has no friction

    F=0 at a smooth contactF=0\text{ at a smooth contact}

    A smooth surface can only push at right angles to itself.

  11. Choose a pivot that removes an unknown force

    take moments about the line of an unknown\text{take moments about the line of an unknown}

    Taking moments through an unknown makes that force contribute no moment.

  12. Resolve the forces into horizontal and vertical components

    components: Fcos⁡θ, Fsin⁡θ\text{components: }F\cos\theta,\ F\sin\theta

    Each force is split along two perpendicular directions before balancing.

  13. Recall that the perpendicular distance uses the sine of the angle

    d=ℓsin⁡θd=\ell\sin\theta

    The moment arm of a vertical force about the foot of an inclined rod is the horizontal offset.

  14. Recall the tangent of an angle from a right triangle

    tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

    A Pythagorean triple fixes the sine and cosine exactly.

  15. State the modelling assumptions

    rigid body, light string, uniform where stated\text{rigid body, light string, uniform where stated}

    These are the standard assumptions behind every statics calculation.

  16. Note that the reaction at a hinge has two components

    R=H2+V2R=\sqrt{H^{2}+V^{2}}

    A hinge (or pin) can push in any direction, so it has a horizontal and a vertical part.

  17. Recall how to combine perpendicular components into a magnitude

    ∣R∣=H2+V2|\mathbf{R}|=\sqrt{H^{2}+V^{2}}

    The resultant of two perpendicular components is found by Pythagoras.

  18. Recall how to find the direction of a resultant

    tan⁡ϕ=VH\tan\phi=\frac{V}{H}

    The angle to the horizontal is the arctangent of the vertical over the horizontal part.

  19. Take the inverse tangent

    θ=38.7∘\theta=38.7^{\circ}

    This is the least angle for which the ladder can rest without slipping.

Answer
θ=38.7∘\theta=38.7^{\circ}
Question 4
9 markschallenging
A uniform ladder ABAB of weight 380380 N rests with its top BB against a smooth vertical wall and its foot AA on rough horizontal ground, the coefficient of friction between the ladder and the ground being 0.450.45. A person of weight 760760 N stands at a point a fraction 35\frac{3}{5} of the way up the ladder from AA. Find the least angle that the ladder can make with the horizontal if it is not to slip. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Resolve vertically

    N=380+760 NN=380+760\ \text{N}

    The normal reaction at the foot balances the total weight.

  2. Set the friction to its limiting value

    F=μN=0.45×1140F=\mu N=0.45\times 1140

    The ladder is about to slip, so the friction is fully mobilised.

  3. Take moments about the foot AA and equate to the friction

    μN=(12×380+760×35)cot⁡θ\mu N=\left(\tfrac{1}{2}\times 380+760\times \frac{3}{5}\right)\cot\theta

    The wall reaction equals the friction and also equals (…)cot⁡θ(\ldots)\cot\theta.

  4. Rearrange for tan⁡θ\tan\theta

    tan⁡θ=646513=1.259\tan\theta=\frac{646}{513}=1.259

    Any steeper angle makes the required friction smaller than μN\mu N.

  5. Take the inverse tangent

    θ=arctan⁡(1.259)\theta=\arctan\left(1.259\right)

    The least angle follows from the inverse tangent of the ratio.

  6. State the conditions for the equilibrium of a rigid body

    ΣFx=0,ΣFy=0,ΣM=0\Sigma F_x=0,\quad\Sigma F_y=0,\quad\Sigma M=0

    A rigid body is in equilibrium when the forces and the moments both balance.

  7. Recall the definition of the moment of a force

    M=FdM=Fd

    The moment is the force multiplied by the perpendicular distance to the pivot.

  8. Recall the law of friction at a rough contact

    F≤μRF\le\mu R

    The friction can take any value up to μR\mu R; at the point of slipping it equals μR\mu R.

  9. Note that the weight of a uniform body acts at its centre

    W acts at the midpointW\text{ acts at the midpoint}

    A uniform rod or lamina has its centre of mass at its geometric centre.

  10. Recall that a smooth contact has no friction

    F=0 at a smooth contactF=0\text{ at a smooth contact}

    A smooth surface can only push at right angles to itself.

  11. Choose a pivot that removes an unknown force

    take moments about the line of an unknown\text{take moments about the line of an unknown}

    Taking moments through an unknown makes that force contribute no moment.

  12. Resolve the forces into horizontal and vertical components

    components: Fcos⁡θ, Fsin⁡θ\text{components: }F\cos\theta,\ F\sin\theta

    Each force is split along two perpendicular directions before balancing.

  13. Recall that the perpendicular distance uses the sine of the angle

    d=ℓsin⁡θd=\ell\sin\theta

    The moment arm of a vertical force about the foot of an inclined rod is the horizontal offset.

  14. Recall the tangent of an angle from a right triangle

    tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

    A Pythagorean triple fixes the sine and cosine exactly.

  15. State the modelling assumptions

    rigid body, light string, uniform where stated\text{rigid body, light string, uniform where stated}

    These are the standard assumptions behind every statics calculation.

  16. Note that the reaction at a hinge has two components

    R=H2+V2R=\sqrt{H^{2}+V^{2}}

    A hinge (or pin) can push in any direction, so it has a horizontal and a vertical part.

  17. State the least angle

    θ=51.5∘\theta=51.5^{\circ}

    Below this angle the ground could not supply enough friction.

Answer
θ=51.5∘\theta=51.5^{\circ}
Question 5
9 markschallenging
A uniform rod ABAB of weight 320320 N is freely hinged at AA to a vertical wall. The rod is held in a horizontal position by a light inextensible string attached to the end BB and to a point of the wall above AA, the string making an angle α\alpha with the rod, where tan⁡α=2021\tan\alpha=\frac{20}{21}. A particle of weight 100100 N hangs from the end BB. Find the angle that the force exerted by the hinge on the rod makes with the horizontal. Give your answer to 33 significant figures.
Show worked solution

Worked solution

  1. Take moments about the hinge AA

    Tsin⁡α⋅L=320⋅12L+100⋅LT\sin\alpha\cdot L=320\cdot\tfrac{1}{2}L+100\cdot L

    The hinge reaction has no moment about AA, so it is the natural pivot.

  2. Solve for the tension

    T=12×320+100sin⁡α=377 NT=\frac{\tfrac{1}{2}\times 320+100}{\sin\alpha}=377\ \text{N}

    The length of the rod cancels, leaving the tension in terms of sin⁡α\sin\alpha.

  3. Resolve horizontally for the hinge reaction

    H=Tcos⁡α=273 NH=T\cos\alpha=273\ \text{N}

    The horizontal hinge component balances the horizontal pull of the string.

  4. Resolve vertically for the hinge reaction

    V=320+100−Tsin⁡α=160 NV=320+100-T\sin\alpha=160\ \text{N}

    The vertical hinge component makes up the rest of the load.

  5. Find the direction of the hinge force

    tan⁡ϕ=VH=160273\tan\phi=\frac{V}{H}=\frac{160}{273}

    The angle to the horizontal is the arctangent of V/HV/H.

  6. Note the sense of the hinge components

    H towards the wall, V upwardsH\text{ towards the wall},\ V\text{ upwards}

    Both components come out positive, confirming the assumed directions.

  7. State the conditions for the equilibrium of a rigid body

    ΣFx=0,ΣFy=0,ΣM=0\Sigma F_x=0,\quad\Sigma F_y=0,\quad\Sigma M=0

    A rigid body is in equilibrium when the forces and the moments both balance.

  8. Recall the definition of the moment of a force

    M=FdM=Fd

    The moment is the force multiplied by the perpendicular distance to the pivot.

  9. Recall the law of friction at a rough contact

    F≤μRF\le\mu R

    The friction can take any value up to μR\mu R; at the point of slipping it equals μR\mu R.

  10. Note that the weight of a uniform body acts at its centre

    W acts at the midpointW\text{ acts at the midpoint}

    A uniform rod or lamina has its centre of mass at its geometric centre.

  11. Recall that a smooth contact has no friction

    F=0 at a smooth contactF=0\text{ at a smooth contact}

    A smooth surface can only push at right angles to itself.

  12. Choose a pivot that removes an unknown force

    take moments about the line of an unknown\text{take moments about the line of an unknown}

    Taking moments through an unknown makes that force contribute no moment.

  13. Resolve the forces into horizontal and vertical components

    components: Fcos⁡θ, Fsin⁡θ\text{components: }F\cos\theta,\ F\sin\theta

    Each force is split along two perpendicular directions before balancing.

  14. Recall that the perpendicular distance uses the sine of the angle

    d=ℓsin⁡θd=\ell\sin\theta

    The moment arm of a vertical force about the foot of an inclined rod is the horizontal offset.

  15. State the answer

    ϕ=30.4∘\phi=30.4^{\circ}

    This is the required quantity.

Answer
ϕ=30.4∘\phi=30.4^{\circ}

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