A block of mass $$$5 \mathrm{~kg}$$$ is held in equilibrium near a vertical wall by two light strings and a horizontal force of magnitude $$$X \mathrm{~N}$$$, as shown in the diagram. The two strings are both inclined at $$$60^{\circ}$$$ to the vertical. Given that $$$X=100$$$, find the tension in the lower string. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w21_qp_42 Year：2021 Question No：6(a)

Answer:

Horizontal: $\quad 100-T_{U} \sin 60-T_{L} \sin 60=0$
Vertical: $\quad T_{U} \cos 60-T_{L} \cos 60-5 g=0$
Perp to $T_{U} \quad T_{L} \cos 30+5 g \cos 30=100 \cos 60$
Solve for either $T_{L}$ or $T_{U}$ using equation(s) with no missing term.
$T_{L}=7.74 \mathrm{~N}$

Knowledge points:

4.1.2 understand the vector nature of force, and find and use components and resultants； Calculations are always required, not approximate solutions by scale drawing.

4.1.3 use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero （Solutions by resolving are usually expected, but equivalent methods (e.g. triangle of forces, Lami’s Theorem, where suitable) are also acceptable; these other methods are not required knowledge, and will not be referred to in questions.）

Answer:

Knowledge points:

Solution:

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