Question

a 300 - kg box rests on a platform attached to a forklift. starting from rest at time = 0, the box is lowered with a downward acceleration of 1.5 m/s². a. determine the upward force exerted by the horizontal platform on the box as it is lowered. at time t = 0, the forklift also begins to move forward with an acceleration of 2 m/s² while lowering the box as described above. the box does not slip or tip over. b. determine the frictional force on the box. c. given that the box does not slip, determine the minimum possible coefficient of friction between the box and the platform. d. determine an equation for the path of the box that expresses y as a function of x (and not of t), assuming that, at time t = 0, the box has a horizontal position x = 0 and a vertical position y = 2 m above the ground, with zero velocity. e. on the axes below sketch the path taken by the box

Explanation:

Step1: Analyze vertical forces for part a

Apply Newton's second - law $F_{net}=ma$. The forces acting on the box vertically are the force of gravity $F_g = mg$ (downward) and the normal force $N$ (upward). The net force $F_{net}=mg - N$. Given $m = 300\ kg$, $g=9.8\ m/s^2$ and $a = 1.5\ m/s^2$ (downward), so $mg - N=ma$. Rearranging for $N$ gives $N=m(g - a)$.
$N=300\times(9.8 - 1.5)=300\times8.3 = 2490\ N$.

Step2: Analyze horizontal forces for part b

In the horizontal direction, the only force causing the acceleration of the box is the frictional force $f$. According to Newton's second - law $F = ma$. Here, $m = 300\ kg$ and $a_x=2\ m/s^2$. So $f=ma_x$.
$f = 300\times2=600\ N$.

Step3: Find minimum coefficient of friction for part c

The maximum static frictional force is given by $f=\mu_sN$. We know $f = 600\ N$ (from part b) and $N = 2490\ N$ (from part a). Rearranging $\mu_s=\frac{f}{N}$ gives $\mu_s=\frac{600}{2490}\approx0.241$.

Step4: Find equations of motion for part d

The equations of motion for the horizontal and vertical directions are:
Horizontal: $x = v_{0x}t+\frac{1}{2}a_xt^2$, with $v_{0x}=0$ and $a_x = 2\ m/s^2$, so $x=\frac{1}{2}\times2t^2=t^2$.
Vertical: $y=y_0+v_{0y}t-\frac{1}{2}a_yt^2$, with $y_0 = 2\ m$, $v_{0y}=0$ and $a_y = 1.5\ m/s^2$, so $y=2-\frac{1}{2}\times1.5t^2=2 - 0.75t^2$.
Eliminating $t$ from the two equations, since $t=\sqrt{x}$, we substitute $t$ into the $y$ - equation: $y = 2-0.75x$.

Step5: Sketch the path for part e

The equation $y = 2-0.75x$ is a straight - line equation with a $y$ - intercept of $y = 2$ and a slope of $- 0.75$. Start at the point $(0,2)$ on the $y$ - axis and draw a straight line with a negative slope of $-0.75$.

Answer:

a. $2490\ N$
b. $600\ N$
c. $0.241$
d. $y = 2-0.75x$
e. Sketch a straight line with $y$ - intercept at $(0,2)$ and slope $-0.75$.