- #1
capn awesome
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Hi. I'm new here, and decided to sign up now that physics is actually difficult for me in first year university... Hope to find the help I need, and help those that I can. Anyway..
I'm having trouble with these wordy physics problems...
1) A light rope passing over a light frictionless pulley is fastened to a platform and a man on the platform holds the other end of the rope. The man pulls on the rope with sufficient force to give himself and the platform an upward acceleration of 0.91m/s^2. Find the tension in the rope and the reaction between the man's feet and platform on which he stands. The mass of the man and the platform are 72kg and 36 kg respectively.
So I tried drawing my FBD, where acceleration points upwards from the platform, and tension points upwards on each side of the pulley. Fg points downwards from the platform. Finally there is an applied force, Fp, opposing tension on the side with the man. I stated that F = ma, added the two masses together to find the force, and divided that by two to get the tension. Am I on the right track...? And then what is it asking me to do for the "reaction between the man's feet and the platform"?
Fp = a(m1+m2)
Ftension = Fp/2
2) A long plank weighing 142N slides on a level frictionless surface with initial speed = 7.3m/s moving the direction of its length. A block weighing 35.5N with initial speed = 0m/s is set down on the plank. If coefficient of kinetic friction between the plank and block is 0.5, what is the acceration of the plank and block?
Also, what is the length of the skid mark made by the block on the plank?
So far, I have found the masses of each using the given weights, found the force of kinetic friction by multiplying the coefficient with Fn. Where would I go with that to find the deceleration of the plank? I assume the masses will have to combine, but what does net force equal in order to solve for acceleration?
mass of plank = m1 = 14.5kg
mass of block = m2 = 3.6kg
kinetic friction = μFn = 17.8N
I'm having trouble with these wordy physics problems...
1) A light rope passing over a light frictionless pulley is fastened to a platform and a man on the platform holds the other end of the rope. The man pulls on the rope with sufficient force to give himself and the platform an upward acceleration of 0.91m/s^2. Find the tension in the rope and the reaction between the man's feet and platform on which he stands. The mass of the man and the platform are 72kg and 36 kg respectively.
So I tried drawing my FBD, where acceleration points upwards from the platform, and tension points upwards on each side of the pulley. Fg points downwards from the platform. Finally there is an applied force, Fp, opposing tension on the side with the man. I stated that F = ma, added the two masses together to find the force, and divided that by two to get the tension. Am I on the right track...? And then what is it asking me to do for the "reaction between the man's feet and the platform"?
Fp = a(m1+m2)
Ftension = Fp/2
2) A long plank weighing 142N slides on a level frictionless surface with initial speed = 7.3m/s moving the direction of its length. A block weighing 35.5N with initial speed = 0m/s is set down on the plank. If coefficient of kinetic friction between the plank and block is 0.5, what is the acceration of the plank and block?
Also, what is the length of the skid mark made by the block on the plank?
So far, I have found the masses of each using the given weights, found the force of kinetic friction by multiplying the coefficient with Fn. Where would I go with that to find the deceleration of the plank? I assume the masses will have to combine, but what does net force equal in order to solve for acceleration?
mass of plank = m1 = 14.5kg
mass of block = m2 = 3.6kg
kinetic friction = μFn = 17.8N