F

To find the force F acting on the block at t = 6.5 sec, we'll utilize the properties of angular momentum (L) and the radius (r) from the center of rotation to the point where the force is applied.
Step 1: Given that the angular momentum L about point O varies with time, we can differentiate L with respect to time to find the torque. The equation for torque (\tau) is given by \( \tau = \frac{dL}{dt} \).
Step 2: The torque can also be expressed in terms of force and distance: \( \tau = r \cdot F \cdot \sin(\beta) \).
Here, r = 6 inches = 0.5 ft, and β = 60º.
Step 3: Calculate the sine component: \( \sin(60º) = \frac{\sqrt{3}}{2} \).
Step 4: Set the two equations for torque equal: \( \frac{dL}{dt} = r \cdot F \cdot \sin(\beta) \).
Step 5: We'll substitute in the known values and solve for F.
Thus, we can rearrange this into \( F = \frac{\frac{dL}{dt}}{r \cdot \sin(60º)} \).
Make sure to compute \( \frac{dL}{dt} \) from the graph provided to find the value of F.