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So now, 6.4-2Fp=ma

and I know the Torque is (Fp+Fg)R they both act in the same direction, hence why I added them together. I know, Torque also equals Iα where α=a/r and I=.5MR^2 so all together .5MR^2(a/R)=(Fp+Fg)(R) where a=2(Fp+Fg)/M. I want to eliminate Fg so I have only two variables to solve for a, a being one of them. I am eluded to there must be another way of doing this that I am not aware of, in which I can solve for a where I do not have a second friction force. I know on a single cylinder, I have mg pointing down, and the normal force pointing up, I have the mg from the plank pointing down onto the cylinder, I have the force of friction from the ground pointing to the right. and I have the force of friction from the plank pointing to the left. As of now I have been measuring torque from the center of mass of the cylinder. If there is any other way of measuring the torque, I'm not aware of it. Is the torque applied from the bottom of the cylinder would the torque be Fp(2R)=(.5*M2R^2)(a/2R)?

and I know the Torque is (Fp+Fg)R they both act in the same direction, hence why I added them together. I know, Torque also equals Iα where α=a/r and I=.5MR^2 so all together .5MR^2(a/R)=(Fp+Fg)(R) where a=2(Fp+Fg)/M. I want to eliminate Fg so I have only two variables to solve for a, a being one of them. I am eluded to there must be another way of doing this that I am not aware of, in which I can solve for a where I do not have a second friction force. I know on a single cylinder, I have mg pointing down, and the normal force pointing up, I have the mg from the plank pointing down onto the cylinder, I have the force of friction from the ground pointing to the right. and I have the force of friction from the plank pointing to the left. As of now I have been measuring torque from the center of mass of the cylinder. If there is any other way of measuring the torque, I'm not aware of it. Is the torque applied from the bottom of the cylinder would the torque be Fp(2R)=(.5*M2R^2)(a/2R)?

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