Solution to Question 26
1. Analyzing Forces along the Line of Fastest Descent
Let the mass of the block be $m$, the angle of inclination be $\theta$, and the coefficient of kinetic friction be $\mu$. The limiting (kinetic) friction force is $f = \mu mg \cos\theta$. The component of gravity along the plane is $mg \sin\theta$.
Let the mass of the block be $m$, the angle of inclination be $\theta$, and the coefficient of kinetic friction be $\mu$. The limiting (kinetic) friction force is $f = \mu mg \cos\theta$. The component of gravity along the plane is $mg \sin\theta$.
Case 1: Dragging Up
To drag the block up at a constant speed, the applied force $F_1$ must overcome both gravity and friction (which acts downwards). $$ F_1 = mg \sin\theta + f \quad \dots(1) $$ Case 2: Dragging Down
To drag the block down at a constant speed, the applied force $F_2$ acts downwards. Friction acts upwards (opposing motion). Since we have to “drag” it, friction must be greater than the gravity component. $$ F_2 + mg \sin\theta = f \implies F_2 = f – mg \sin\theta \quad \dots(2) $$
To drag the block up at a constant speed, the applied force $F_1$ must overcome both gravity and friction (which acts downwards). $$ F_1 = mg \sin\theta + f \quad \dots(1) $$ Case 2: Dragging Down
To drag the block down at a constant speed, the applied force $F_2$ acts downwards. Friction acts upwards (opposing motion). Since we have to “drag” it, friction must be greater than the gravity component. $$ F_2 + mg \sin\theta = f \implies F_2 = f – mg \sin\theta \quad \dots(2) $$
2. Expressing $f$ and $mg\sin\theta$ in terms of $F_1$ and $F_2$
Adding equations (1) and (2): $$ F_1 + F_2 = 2f \implies f = \frac{F_1 + F_2}{2} $$ Subtracting equation (2) from (1): $$ F_1 – F_2 = 2mg \sin\theta \implies mg \sin\theta = \frac{F_1 – F_2}{2} $$
Adding equations (1) and (2): $$ F_1 + F_2 = 2f \implies f = \frac{F_1 + F_2}{2} $$ Subtracting equation (2) from (1): $$ F_1 – F_2 = 2mg \sin\theta \implies mg \sin\theta = \frac{F_1 – F_2}{2} $$
3. Case 3: Minimum Horizontal Force $F$
Now, a horizontal force $F$ is applied parallel to the plane. The block is on the verge of sliding. The forces acting in the plane of the incline are:
Now, a horizontal force $F$ is applied parallel to the plane. The block is on the verge of sliding. The forces acting in the plane of the incline are:
- The component of gravity $mg \sin\theta$ (acting down the slope).
- The applied force $F$ (acting horizontally, perpendicular to gravity component).
4. Solution
Substitute the values from step 2: $$ F^2 + \left( \frac{F_1 – F_2}{2} \right)^2 = \left( \frac{F_1 + F_2}{2} \right)^2 $$ $$ F^2 = \frac{(F_1 + F_2)^2}{4} – \frac{(F_1 – F_2)^2}{4} $$ Using the identity $(a+b)^2 – (a-b)^2 = 4ab$: $$ F^2 = \frac{4 F_1 F_2}{4} = F_1 F_2 $$ $$ F = \sqrt{F_1 F_2} $$
Substitute the values from step 2: $$ F^2 + \left( \frac{F_1 – F_2}{2} \right)^2 = \left( \frac{F_1 + F_2}{2} \right)^2 $$ $$ F^2 = \frac{(F_1 + F_2)^2}{4} – \frac{(F_1 – F_2)^2}{4} $$ Using the identity $(a+b)^2 – (a-b)^2 = 4ab$: $$ F^2 = \frac{4 F_1 F_2}{4} = F_1 F_2 $$ $$ F = \sqrt{F_1 F_2} $$
Answer: The minimum horizontal force required is $F = \sqrt{F_1 F_2}$.