Question

A horizontal forces F pushes a 9kg load up an inclined plane with an acceleration of 1.2m/s^2. what is the magnitude of the force if the plane makes an angle of 30° with the horizontal and the coefficient of kinetic friction between the load and the plane is 0.15?

Answer 1

Let's apply the Newton's Second Law of Motion in projections on axis $x$- and $y$:

$$\sum F_x = ma_x = ma,$$ $$\sum F_y = ma_y = 0,$$ $$F_{appl} - mgsin\theta - F_{fr} = ma,$$ $$N - mgcos\theta = 0,$$ $$F_{appl} - mgsin\theta - \mu_k N = ma,$$ $$N = mgcos\theta,$$ $$F_{appl} - mgsin\theta - \mu_k mgcos\theta = ma.$$

From this equation we can find the the magnitude of the applied force:

$$F_{appl} = ma + mg(sin\theta + \mu_k cos\theta),$$

$F_{appl} = 9 \ kg \cdot 1.2 \ \dfrac{m}{s^2} + 9 \ kg \cdot 9.8 \ \dfrac{m}{s^2} \cdot (sin30^{\circ} + 0.15 \cdot cos30^{\circ}) = 66.3 \ N.$

Answer:

$F_{appl} = 66.3 \ N.$