Answer to Question #150281 in Physics for Khan Asf

Question

There is a block on a horizontal surface. Coefficient of friction between the bar and the surface 0.2. If a force F is applied to the bar, directed upwards at an angle of 30° to the horizon, then the bar will move uniformly and rectilinearly along the table. Find the acceleration of the bar if a force of 1.4 N is applied to it in the same direction. Take g = 10 m/s^2.

Answer 1

If the block moves without acceleration in the first case, write equilibrium of forces:

$$F\text{ cos}30°-f=0,\\ N+F\text{ sin}30°-mg=0.\\ N=mg-F\text{ sin}30°,\\ f=\mu N=\mu(mg-F\text{ sin}30°).\\\space\\ F\text{ cos}30°=\mu mg-\mu F\text{ sin}30°,\\\space\\ F=\frac{\mu mg}{\text{cos}30°+\mu \text{sin}30°}.$$

Now, add 1.4 N to our force F:

$$(F+1.4)\text{ cos}30°-f=ma,\\ N+(F+1.4)\text{ sin}30°-mg=0.\\ N=mg-(F+1.4)\text{ sin}30°,\\ f=\mu N=\mu(mg-(F+1.4)\text{ sin}30°).\\\space\\ a=\frac{(F+1.4)\text{ cos}30°-f}{m},\\\space\\ a=\frac{F\text{ cos}30°+1.4\text{ cos}30°-\mu mg-\mu F\text{ sin}30°-1.4\mu \text{ sin}30°)}{m},\\\space\\ a=\frac{F(\text{cos}30°-\mu \text{ sin}30°)+1.4(\text{cos}30°-\mu\text{ sin}30°)-\mu mg}{m},\\\space\\ a=\frac{F(\text{cos}30°-\mu \text{ sin}30°)+1.4(\text{cos}30°-\mu\text{ sin}30°)}{m}-\mu g.$$

Substitute F we found previously:

$$a=\mu g\bigg(\frac{\text{cos}30°-\mu \text{ sin}30°}{\text{cos}30°+\mu \text{sin}30°}-1\bigg)+\frac{1.4(\text{cos}30°-\mu\text{ sin}30°)}{m}.$$

With given numerical values, the answer is

$$a=\bigg[1.59+\frac{1.07}{m}\bigg]\text{ m/s}^2.$$