The figure below shows a toolbox of weight 300N on a rough horizontal plane having a coefficient of friction as 0.2. Calculate the magnitude of the force which can pull the box while acting at an angle of 25 degrees with the horizontal.

The weight of the toolbox, $W = 300 \, \text{N}$ (taken from the diagram, as it is more clearly indicated than the text).
The coefficient of friction, $\mu = 0.2$.
The angle of the pulling force with the horizontal, $\theta = 25^\circ$.

Let $P$ be the magnitude of the pulling force and $R$ be the normal force exerted by the surface on the toolbox. The friction force $f_f$ opposes the motion.
The friction force is given by:

$$f_f = \mu R$$

For vertical equilibrium (no vertical acceleration):
The sum of upward forces equals the sum of downward forces.
The upward forces are the normal force $R$ and the vertical component of the pulling force $P \sin \theta$.
The downward force is the weight $W$.

$$R + P \sin \theta = W$$ $$R = W - P \sin \theta \quad (1)$$

For horizontal equilibrium (constant velocity or on the verge of motion):
The sum of forces in the direction of motion equals the sum of forces opposing motion.
The force in the direction of motion is the horizontal component of the pulling force $P \cos \theta$.
The force opposing motion is the friction force $f_f$.

$$P \cos \theta = f_f \quad (2)$$

Substitute $f_f = \mu R$ into equation (2):

$$P \cos \theta = \mu R$$

$$P \cos \theta = \mu (W - P \sin \theta)$$ $$P \cos \theta = \mu W - \mu P \sin \theta$$

Move all terms containing $P$ to one side:

$$P \cos \theta + \mu P \sin \theta = \mu W$$

Factor out $P$:

$$P (\cos \theta + \mu \sin \theta) = \mu W$$

Solve for $P$:

$$P = \frac{\mu W}{\cos \theta + \mu \sin \theta}$$

Given $W = 300 \, \text{N}$, $\mu = 0.2$, and $\theta = 25^\circ$.
First, calculate $\cos 25^\circ$ and $\sin 25^\circ$:
$\cos 25^\circ \approx 0.9063$
$\sin 25^\circ \approx 0.4226$

Now substitute these values into the equation for $P$:

$$P = \frac{0.2 \times 300}{0.9063 + 0.2 \times 0.4226}$$ $$P = \frac{60}{0.9063 + 0.08452}$$ $$P = \frac{60}{0.99082}$$ $$P \approx 60.5555 \, \text{N}$$

Rounding to three significant figures:

$$P \approx 60.6 \, \text{N}$$

The magnitude of the force required to pull the box is $\text{60.6 N}$.