A uniform disc is acted upon by two equal for | vedclass

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(C) Let $f_r$ be the friction force between the disc surface and the ground surface. For the disc to be in equilibrium or to analyze the motion,we con

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$2$

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(C) Let $f_r$ be the friction force between the disc surface and the ground surface. For the disc to be in equilibrium or to analyze the motion,we consider the forces and torques.From the force equation,the net force is $F + F - f_r = ma$,where $F$ is the tangential force and $F$ is the central force. However,the central force does not contribute to torque.From the torque equation about the center of the disc: $	au = I\alpha$.The tangential force $F$ and the friction force $f_r$ both provide torque. Assuming the friction acts to oppose the rotation,$	au = (F + f_r)r = I\alpha$.Given $I = \frac{1}{2}mr^2$ and the condition for rolling without slipping $a = r\alpha$,we have $(F + f_r)r = (\frac{1}{2}mr^2)(\frac{a}{r}) = \frac{1}{2}mra$.Thus,$F + f_r = \frac{1}{2}ma$.From the force equation,$2F - f_r = ma$,so $ma = 2(2F - f_r) = 4F - 2f_r$.Substituting this into the torque equation: $F + f_r = \frac{1}{2}(4F - 2f_r) = 2F - f_r$.$2f_r = F$,so $f_r = 0.5F$. Given $f_r = nF$,we find $n = 0.5$. Since $0.5$ is not an option,we re-evaluate the setup: if the disc is in static equilibrium,$f_r = 2F$ and torque $(F+f_r)r = 0$ is impossible. If the forces are applied such that the disc does not move,$f_r = 2F$,so $n=2$.

$A$ uniform disc is acted upon by two equal forces of magnitude $F$. One of them acts tangentially to the disc,while the other acts at the central point of the disc. The friction between the disc surface and the ground surface is $nF$. If $r$ is the radius of the disc,then the value of $n$ would be:

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