Question

A block of mass $\sqrt{2}$ kg is placed on a rough horizontal surface. A force 'F' acting upwards at an angle of 45$^\circ$ with the horizontal causes the block to start motion. If the coefficient of static friction between the surface and the block is 0.25, the magnitude of the force 'F' is (Acceleration due to gravity = 10 ms$^{-2}$)

• A. 0.5 N
• B. 2 N
• C. 4 N
• D. 8 N

Hint:

When a force is applied at an angle to a body on a horizontal surface, it affects both the driving force (horizontal component) and the normal force (vertical component). An upward component of the force reduces the normal force and hence reduces the maximum possible friction.

Answer 1

The Correct Option is C

Step 1: Draw a free-body diagram and resolve the forces.
The forces acting on the block are: 1. Weight $W = mg$ acting downwards. 2. Normal force $N$ acting upwards. 3. Applied force $F$ acting at $45^\circ$ to the horizontal. Its components are $F\cos(45^\circ)$ horizontally and $F\sin(45^\circ)$ vertically. 4. Static friction force $f_s$ acting horizontally, opposing motion. When the block is about to start moving, the friction is at its maximum value, $f_s = \mu_s N$.

Step 2: Apply the conditions for equilibrium.
Vertical equilibrium: The net vertical force is zero. \[ N + F\sin(45^\circ) - mg = 0 \implies N = mg - F\sin(45^\circ). \] Horizontal equilibrium: The horizontal component of F is balanced by the maximum static friction. \[ F\cos(45^\circ) = f_s = \mu_s N. \]

Step 3: Substitute and solve for F.
Substitute the expression for $N$ into the horizontal equation: \[ F\cos(45^\circ) = \mu_s (mg - F\sin(45^\circ)). \] We know $\sin(45^\circ) = \cos(45^\circ) = 1/\sqrt{2}$. Let's use this value. \[ \frac{F}{\sqrt{2}} = \mu_s \left(mg - \frac{F}{\sqrt{2}}\right). \] Rearrange to solve for F: \[ \frac{F}{\sqrt{2}} = \mu_s mg - \mu_s \frac{F}{\sqrt{2}}. \] \[ \frac{F}{\sqrt{2}}(1+\mu_s) = \mu_s mg. \] \[ F = \frac{\mu_s mg \sqrt{2}}{1+\mu_s}. \]

Step 4: Substitute the numerical values.
Given $m=\sqrt{2}$ kg, $g=10$ m/s$^2$, $\mu_s = 0.25 = 1/4$. \[ F = \frac{(1/4) (\sqrt{2})(10) \sqrt{2}}{1+1/4} = \frac{(1/4)(20)}{5/4}. \] \[ F = \frac{5}{5/4} = 5 \times \frac{4}{5} = 4 \text{ N}. \] \[ \boxed{F = 4 \text{ N}}. \]