A rectangular box is sliding on a smooth inclined plane of inclination α as shown in the figure. At t = 0, the box starts to move on the inclined plane. A bolt starts to fall from point A. Find the time after which bolt strikes the bottom surface of the box.

Answer

To solve the problem of determining the time after which the bolt strikes the bottom surface of the box sliding down an inclined plane, we can follow these steps:

### Step 1: Understand the Motion of the Box and the Bolt
At time \( t = 0 \), the box starts moving down the inclined plane with an angle of inclination \( \alpha \). Simultaneously, a bolt starts to fall from point A. The box and the bolt will have the same horizontal acceleration due to the smooth incline.

### Step 2: Identify the Acceleration of the Bolt
The acceleration of the box down the incline is given by the component of gravitational acceleration acting along the incline:
\[ a_{\text{box}} = g \sin \alpha \]
However, since the bolt falls vertically, its vertical acceleration is simply \( g \). The bolt will also have a horizontal component of motion due to the incline:
\[ a_{\text{bolt}} = g \cos \alpha \]

### Step 3: Set Up the Equation of Motion for the Bolt
The bolt starts falling from rest, so its initial velocity \( u = 0 \). The distance \( s \) that the bolt falls vertically is equal to the distance \( l \) that the box slides down the incline. We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ l = 0 + \frac{1}{2} (g \cos \alpha) t^2 \]

### Step 4: Rearrange the Equation to Solve for Time \( t \)
Rearranging the equation gives:
\[ l = \frac{1}{2} g \cos \alpha \cdot t^2 \]
Multiplying both sides by 2:
\[ 2l = g \cos \alpha \cdot t^2 \]
Now, solving for \( t^2 \):
\[ t^2 = \frac{2l}{g \cos \alpha} \]
Taking the square root:
\[ t = \sqrt{\frac{2l}{g \cos \alpha}} \]

### Step 5: Conclusion
The time after which the bolt strikes the bottom surface of the box is given by:
\[ t = \sqrt{\frac{2l}{g \cos \alpha}} \]