Let's analyze the forces acting on the box:
The component of weight parallel to the incline is mg sin30° = 50 × $\frac{1}{2}$ = 25 N (down the incline).
Applying Newton's second law along the incline (upward direction as positive):
T − mg sin 30° = ma
30 N − 25 N = 5 kg × a
a = $\frac{5}{5}$ = 1 m s-2
A body of mass 1kg is suspended with the help of two strings making angles as shown in the figure. Magnitude of tensions $ T_1 $ and $ T_2 $, respectively, are (in N):
Consider a water tank shown in the figure. It has one wall at \(x = L\) and can be taken to be very wide in the z direction. When filled with a liquid of surface tension \(S\) and density \( \rho \), the liquid surface makes angle \( \theta_0 \) (\( \theta_0 < < 1 \)) with the x-axis at \(x = L\). If \(y(x)\) is the height of the surface then the equation for \(y(x)\) is: (take \(g\) as the acceleration due to gravity)