More practice questions

  1. Q4. A weight of 30 N hangs from a knot held by two strings. The three strings at the knot make angles of 120° with one another. Use Lami’s theorem to find the tension in each upper string.

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    Each force is divided by the sine of the angle between the other two, and every such angle is 120°:

    T1sin⁡120∘=T2sin⁡120∘=30sin⁡120∘\frac{T_1}{\sin 120^\circ} = \frac{T_2}{\sin 120^\circ} = \frac{30}{\sin 120^\circ}

    So T1 = T2 = 30 N.

  2. Q5. Two forces act on a particle (in newtons):

    F⃗1=2i^+3j^\vec F_1 = 2\hat i + 3\hat j F⃗2=−5i^+j^\vec F_2 = -5\hat i + \hat j

    Find the third force that keeps it in equilibrium, and its magnitude.

    Show answerHide answer F⃗3=−(F⃗1+F⃗2)=−(−3i^+4j^)=(3i^−4j^) N\begin{aligned} \vec F_3 &= -(\vec F_1 + \vec F_2) \\ &= -(-3\hat i + 4\hat j) \\ &= (3\hat i - 4\hat j)\ \text{N} \end{aligned}

    Its magnitude is 32+42\sqrt{3^{2} + 4^{2}}, that is 5 N.

  3. Q6. A 1 kg lamp hangs at the middle of a wire. The wire sags so that each half makes 5° with the horizontal. Find the tension in the wire, and explain why the wire can never be pulled perfectly straight. Take g = 9.8 m s−2.

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    Vertically, 2T sin 5° = mg, so

    T=9.82×0.08716≈56.2 N\begin{aligned} T &= \frac{9.8}{2 \times 0.08716} \\ &\approx 56.2\ \text{N} \end{aligned}

    almost six times the lamp’s weight. As the angle tends to zero, sin θ → 0 and T = mg/(2 sin θ) grows without limit, so no finite tension can hold the load with a perfectly straight wire.

  4. Q7. A picture of weight 20 N hangs from a nail by a string whose two halves make 60° with the vertical on either side. Find the tension in the string.

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    For the symmetric hanging arrangement, horizontal balance gives equal tensions T in the two halves. Vertically, 2T cos 60° = 20, so

    T=202×0.5=20 NT = \frac{20}{2 \times 0.5} = 20\ \text{N}