An object of weight 500 N is pulled up a plane inclined at 15° using an effort of 750 N as shown in figure Q1a. Determine the coefficient of friction between the object and the plane.

The normal force ($N$) acts perpendicular to the plane, balancing the perpendicular component of the weight ($W \cos \theta$).

$$N = W \cos \theta$$ $$N = 50 \, \text{N} \times \cos 10^\circ$$ $$N \approx 50 \, \text{N} \times 0.9848$$ $$N \approx 49.24 \, \text{N}$$

The frictional force ($F_f$) opposes the direction of motion. Since the block is being dragged down the incline, the frictional force acts up the incline.

$$F_f = \mu N$$ $$F_f = 0.2 \times 49.24 \, \text{N}$$ $$F_f \approx 9.848 \, \text{N}$$

For the block to be dragged down at a constant velocity (or at the point of impending motion), the forces parallel to the plane must be balanced.
The forces acting down the incline are the applied force ($P$) and the component of weight parallel to the plane ($W \sin \theta$).
The force acting up the incline is the frictional force ($F_f$).

$$P + W \sin \theta = F_f$$

Rearrange the equation from Step 3:

$$P = F_f - W \sin \theta$$

Substitute $F_f = \mu N = \mu W \cos \theta$:

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

Factor out $W$:

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

Now, substitute the given values:

$$P = 50 \, \text{N} \times (0.2 \times \cos 10^\circ - \sin 10^\circ)$$ $$P = 50 \, \text{N} \times (0.2 \times 0.98480775 - 0.173648177)$$ $$P = 50 \, \text{N} \times (0.19696155 - 0.173648177)$$ $$P = 50 \, \text{N} \times (0.023313373)$$ $$P \approx 1.16566865 \, \text{N}$$

Rounding to three significant figures:

$$P \approx 1.17 \, \text{N}$$

The force required to drag the block down the gradient is $1.17 \, \text{N}$.