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(A) Let the force $\overrightarrow{P}$ be applied at point $A$ in the downward vertical direction. We need to resolve this force along the directions

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(A) Let the force $\overrightarrow{P}$ be applied at point $A$ in the downward vertical direction. We need to resolve this force along the directions of the rods $AB$ and $AC$.Let $\overrightarrow{F_{AB}}$ and $\overrightarrow{F_{AC}}$ be the components of force $\overrightarrow{P}$ along $AB$ and $AC$ respectively.According to the parallelogram law of vector addition,$\overrightarrow{P} = \overrightarrow{F_{AB}} + \overrightarrow{F_{AC}}$.Using the Lami's theorem or the sine rule in the triangle formed by the force vectors:The angle between the vertical force $\overrightarrow{P}$ and rod $AB$ is $35^{\circ}$ (as the angle between the vertical and the wall is $90^{\circ}$,and the angle between $AB$ and the wall is $70^{\circ}$,so the angle between $AB$ and the vertical is $90^{\circ} - 70^{\circ} = 20^{\circ}$ is incorrect based on the provided image geometry; let's use the provided diagram).From the diagram,the angle between the vertical force $\overrightarrow{P}$ and rod $AC$ is $35^{\circ}$.The component of force $\overrightarrow{P}$ along the direction of rod $AC$ is given by $P \cos(	heta)$,where $	heta$ is the angle between the force vector and the rod.Thus,the magnitude of the component along $AC$ is $x = 100 \cos(35^{\circ}) \; N$.Given $\cos(35^{\circ}) = 0.819$.$x = 100 	imes 0.819 = 81.9 \; N$.Rounding to the nearest integer,we get $x = 82$.

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