Question

Consider a frame that is made up of two thin massless rods $AB$ and $AC$ as shown in the figure. A vertical force $\overrightarrow{P}$ of magnitude $100N$ is applied at point $A$ of the frame. Suppose the force is $\overrightarrow{P}$ resolved parallel to the arms $AB$ and $AC$ of the frame. The magnitude of the resolved component along the arm $AC$ is $xN$.

[16 March, 2021 (Shift-I)]

The value of $x$, to the nearest integer, is ______ .

[Given : $sin(35^\circ) = 0.573, cos(35^\circ) = 0.819$ $sin(110^\circ) = 0.939, cos(110^\circ) = -0.342$]

Answer 1

42

Answer 2

The problem asks us to find the magnitude of the resolved component of a vertical force $\overrightarrow{P}$ along the arm $AC$ of a frame. The force $\overrightarrow{P}$ has a magnitude of $100N$. The force is resolved parallel to the arms $AB$ and $AC$, meaning $\overrightarrow{P}$ is the resultant of two component forces, $\overrightarrow{F}_{AB}$ along $AB$ and $\overrightarrow{F}_{AC}$ along $AC$.

We need to determine the angles involved in the force triangle or parallelogram. Let's consider a vertical line passing through point $A$. The force $\overrightarrow{P}$ acts along this line, downwards.

1. Angle between $\overrightarrow{P}$ and arm $AB$: The figure shows that arm $AB$ makes an angle of $70^\circ$ with the vertical wall. Assuming the wall is perfectly vertical, the angle between arm $AB$ and the horizontal line (perpendicular to the wall) is $90^\circ - 70^\circ = 20^\circ$. Since $\overrightarrow{P}$ is vertically downwards, the angle between $\overrightarrow{P}$ and arm $AB$ is $20^\circ$. Let this be $\theta_1 = 20^\circ$.

2. Angle between $\overrightarrow{P}$ and arm $AC$: The figure shows that arm $AC$ makes an angle of $145^\circ$ with the vertical wall. This angle is measured from the upward vertical along the wall, clockwise to $AC$. The angle from the upward vertical to the downward vertical is $180^\circ$. Therefore, the angle between arm $AC$ and the downward vertical direction (direction of $\overrightarrow{P}$) is $180^\circ - 145^\circ = 35^\circ$. Let this be $\theta_2 = 35^\circ$. Note that arm $AB$ is to one side of the vertical line through $A$, and arm $AC$ is to the other side.

3. Angle between arms $AB$ and $AC$: Since $AB$ makes an angle of $20^\circ$ with the downward vertical and $AC$ makes an angle of $35^\circ$ with the downward vertical, and they are on opposite sides of the vertical, the angle between $AB$ and $AC$ (i.e., $\angle BAC$) is $\theta_1 + \theta_2 = 20^\circ + 35^\circ = 55^\circ$.

4. Applying the Law of Sines (Lami's Theorem concept for resultant): We have $\overrightarrow{P} = \overrightarrow{F}_{AB} + \overrightarrow{F}_{AC}$. We can form a triangle of forces with sides $P$, $F_{AB}$, and $F_{AC}$. Let's consider the triangle formed by vectors $\overrightarrow{P}$, $\overrightarrow{F}_{AC}$, and a vector parallel to $\overrightarrow{F}_{AB}$ (let's call it $\overrightarrow{F'}_{AB}$) such that $\overrightarrow{F}_{AC} + \overrightarrow{F'}_{AB} = \overrightarrow{P}$. The angles in this force triangle are:

   • Angle opposite to $F_{AC}$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{F'}_{AB}$. Since $\overrightarrow{F'}_{AB}$ is parallel to $\overrightarrow{F}_{AB}$, this angle is $\theta_1 = 20^\circ$.
   • Angle opposite to $F_{AB}$ (or $F'_{AB}$) is the angle between $\overrightarrow{P}$ and $\overrightarrow{F}_{AC}$. This angle is $\theta_2 = 35^\circ$.
   • Angle opposite to $P$ is the angle between $\overrightarrow{F}_{AC}$ and $\overrightarrow{F'}_{AB}$. This is the angle between the two component forces. In the parallelogram formed by $\overrightarrow{F}_{AB}$ and $\overrightarrow{F}_{AC}$, the angle between them is $55^\circ$. The angle inside the force triangle (which is the angle supplementary to the angle between the components when they originate from the same point) is $180^\circ - 55^\circ = 125^\circ$.

   Now apply the Law of Sines: $\frac{P}{\sin(125^\circ)} = \frac{F_{AC}}{\sin(20^\circ)} = \frac{F_{AB}}{\sin(35^\circ)}$ We need to find $F_{AC}$ (which is $x$). $x = F_{AC} = P \frac{\sin(20^\circ)}{\sin(125^\circ)}$ Given $P = 100N$. We know that $\sin(125^\circ) = \sin(180^\circ - 55^\circ) = \sin(55^\circ)$. We are given $\cos(35^\circ) = 0.819$. Since $\sin(55^\circ) = \sin(90^\circ - 35^\circ) = \cos(35^\circ)$, we have $\sin(55^\circ) = 0.819$. We are given $\cos(110^\circ) = -0.342$. Since $\cos(110^\circ) = \cos(90^\circ + 20^\circ) = -\sin(20^\circ)$, we have $-\sin(20^\circ) = -0.342$, which means $\sin(20^\circ) = 0.342$.

   Substitute the values: $x = 100 \times \frac{0.342}{0.819}$ $x = 100 \times 0.4175824...$ $x \approx 41.758$ To the nearest integer, $x = 42$.