A mass of 10kg is hanged on a string. When the string is pulled to a side by 50N horizontal force, the particle is on equilibrium. Find the tension on the string and the angle the... A mass of 10kg is hanged on a string. When the string is pulled to a side by 50N horizontal force, the particle is on equilibrium. Find the tension on the string and the angle the string forms with the vertical. Read more

Steps to Solve

1. Draw a free body diagram

Draw a diagram showing all the forces acting on the mass. These forces include: - Weight of the block ($W = mg$) acting downwards - Tension in the string ($T$) acting along the string - Applied horizontal force ($F$)

2. Resolve the tension $T$ into horizontal and vertical components

The tension $T$ can be resolved into two components: - $T_x = T \sin{\theta}$ (horizontal component) - $T_y = T \cos{\theta}$ (vertical component) Where $\theta$ is the angle the string makes with the vertical.

3. Apply equilibrium conditions

Since the block is in equilibrium, the net force in both the horizontal and vertical directions must be zero. This gives us two equations: - Horizontal equilibrium: $T_x = F \implies T \sin{\theta} = F$ - Vertical equilibrium: $T_y = W \implies T \cos{\theta} = mg$

4. Solve for $\theta$

Divide the horizontal equilibrium equation by the vertical equilibrium equation to eliminate $T$: $$ \frac{T \sin{\theta}}{T \cos{\theta}} = \frac{F}{mg} $$ $$ \tan{\theta} = \frac{F}{mg} $$ $$ \theta = \arctan{\frac{F}{mg}} $$ Substitute the given values $F = 7.5 N$, $m = 2 kg$, and $g = 9.8 m/s^2$: $$ \theta = \arctan{\frac{7.5}{2 \times 9.8}} = \arctan{\frac{7.5}{19.6}} \approx \arctan{0.3827} \approx 20.94^\circ $$

5. Solve for $T$

Using the vertical equilibrium equation $T \cos{\theta} = mg$, we can solve for $T$: $$ T = \frac{mg}{\cos{\theta}} $$ Substitute the values $m = 2 kg$, $g = 9.8 m/s^2$, and $\theta \approx 20.94^\circ$: $$ T = \frac{2 \times 9.8}{\cos{20.94^\circ}} \approx \frac{19.6}{0.9336} \approx 20.99 N $$