Solve the problems and explain in English

Okay Nike🙃, let's tackle question 3. This problem involves forces in equilibrium, so we'll use the conditions that the net force in both horizontal and vertical directions is zero. We'll analyze the forces at points Q and R separately.

Given:

• Tension in horizontal string QR = $30 \text{ N}$
• Angle of string PQ with vertical = $60^\circ$
• Mass attached at R = $2.5 \text{ kg}$
• Acceleration due to gravity $g = 9.81 m/s^2$

Step 1: Analyze forces at point Q. At point Q, the system is in equilibrium. The forces acting are:

1. Tension $T_{PQ}$ (let's call it $T_1$) acting upwards and to the left, at $60^\circ$ to the vertical.
2. Tension $T_{QR}$ acting horizontally to the right, which is $30 \text{ N}$.
3. Weight of mass $m$, $W_m = mg$, acting downwards.

Resolve forces into horizontal and vertical components:

• Horizontal equilibrium: The horizontal component of $T_1$ must balance the tension $T_{QR}$. $T_1 \sin(60^\circ) = 30 N$
• Vertical equilibrium: The vertical component of $T_1$ must balance the weight $W_m$. $T_1 \cos(60^\circ) = mg$

Step 2: Calculate $T_1$ and the mass $m$. From the horizontal equilibrium equation: $T_1 \frac{\sqrt{3}}{2} = 30 N$ $T_1 = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} N$ Now, substitute $T_1$ into the vertical equilibrium equation: $mg = (20\sqrt{3} N) \cos(60^\circ)$ $mg = (20\sqrt{3} N) \times \frac{1}{2}$ $mg = 10\sqrt{3} N$ To find $m$, divide by $g$: $m = \frac{10\sqrt{3}}{9.81} kg$ $m \approx \frac{10 \times 1.73205}{9.81} \approx 1.76559 kg$

a) The mass $m$ Rounding to three significant figures: $m = 1.77 \text{ kg}$

Step 3: Analyze forces at point R. At point R, the system is also in equilibrium. The forces acting are:

1. Tension $T_{RS}$ (let's call it $T$) acting upwards and to the left, at an angle $\theta$ to the vertical.
2. Tension $T_{RQ}$ acting horizontally to the left, which is $30 \text{ N}$ (same tension as $T_{QR}$ but in the opposite direction).
3. Weight of the $2.5 \text{ kg}$ mass, $W_{2.5kg} = (2.5 kg)g$, acting downwards.

Resolve forces into horizontal and vertical components:

• Horizontal equilibrium: The horizontal component of $T$ must balance the tension $T_{RQ}$. $T \sin(\theta) = 30 N \quad (Equation 1)$
• Vertical equilibrium: The vertical component of $T$ must balance the weight $W_{2.5kg}$. $T \cos(\theta) = (2.5 kg)(9.81 m/s^2)$ $T \cos(\theta) = 24.525 N \quad (Equation 2)$

Step 4: Calculate the tension $T$ and the angle $\theta$. To find $T$, square Equation 1 and Equation 2 and add them: $(T \sin(\theta))^2 + (T \cos(\theta))^2 = (30)^2 + (24.525)^2$ $T^2 (\sin^2(\theta) + \cos^2(\theta)) = 900 + 601.475625$ Since $\sin^2(\theta) + \cos^2(\theta) = 1$: $T^2 = 1501.475625$ $T = \sqrt{1501.475625} \approx 38.75 N$

b) The tension $T$ Rounding to three significant figures: $T = 38.8 \text{ N}$

To find $\theta$, divide Equation 1 by Equation 2: $\frac{T \sin(\theta)}{T \cos(\theta)} = \frac{30 N}{24.525 N}$ $\tan(\theta) = \frac{30}{24.525} \approx 1.22324$ $\theta = \arctan(1.22324) \approx 50.73^\circ$

c) The angle $\theta$ Rounding to one decimal place: $\theta = 50.7^\circ$