Question

m2. a 300 - kg box rests on a platform attached to a forklift, shown above. starting from rest at time = 0, the box is lowered with a downward acceleration of 1.5m/s². work m2 on your own paper
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 ( 2m/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 = 2m ) 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)

Using Newton's second law \(F_{net}=ma\). The net force on the box is \(mg - N=ma\) (where \(m = 300\space kg\), \(g = 9.8\space m/s^{2}\), \(a=1.5\space m/s^{2}\)). Rearranging for \(N\) (the normal force, which is the upward force exerted by the platform), we get \(N=m(g - a)\).

$$N = 300\times(9.8 - 1.5)=300\times8.3 = 2490\space N$$

Step2: Analyze horizontal forces (for part b)

In the horizontal direction, using \(F_{net}=ma\). The only horizontal force is the frictional force \(f\). Given \(a_x = 2\space m/s^{2}\) and \(m = 300\space kg\). By \(F = ma\), \(f=ma_x\)

$$f=300\times2=600\space N$$

Step3: Relate friction to normal force (for part c)

The maximum static - friction force is \(f=\mu_sN\). We know \(f = 600\space N\) and \(N = 2490\space N\) from part (a). Rearranging \(\mu_s=\frac{f}{N}\)

$$\mu_s=\frac{600}{2490}=\frac{20}{83}\approx0.24$$

Step4: Find equations of motion (for part d)

For the vertical motion (using \(y - y_0=v_{0y}t+\frac{1}{2}a_yt^{2}\)), with \(y_0 = 2\space m\), \(v_{0y}=0\), \(a_y=1.5\space m/s^{2}\), we have \(y=2-\frac{1}{2}\times1.5t^{2}=2 - 0.75t^{2}\). For the horizontal motion (using \(x - x_0=v_{0x}t+\frac{1}{2}a_xt^{2}\)), with \(x_0 = 0\), \(v_{0x}=0\), \(a_x = 2\space m/s^{2}\), we have \(x=\frac{1}{2}\times2t^{2}=t^{2}\). Solving \(t^{2}=x\) and substituting into the \(y\) - equation: \(y = 2-0.75x\)

Answer:

a. \(2490\space N\)
b. \(600\space N\)
c. \(\frac{20}{83}\approx0.24\)
d. \(y = 2-0.75x\)
e. The path is a straight line. Plot the points by choosing values of \(x\) (e.g., \(x = 0,y = 2\); \(x = 1,y=2 - 0.75\times1=1.25\); \(x = 2,y=2-0.75\times 2 = 0.5\)) and draw a straight - line through them.