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 one year ago
A block of mass m1 = 25 kg rests on a wedge of angle θ = 49∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 7 kg. μk = 0.5. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. g= 9.81.The system is released from rest as shown above, at t = 0.
(a) Find the magnitude of the acceleration of block 1 when it is released?
(b) How many cm down the plane will block 1 have traveled when 0.72 s has elapsed?
 one year ago
A block of mass m1 = 25 kg rests on a wedge of angle θ = 49∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 7 kg. μk = 0.5. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. g= 9.81.The system is released from rest as shown above, at t = 0. (a) Find the magnitude of the acceleration of block 1 when it is released? (b) How many cm down the plane will block 1 have traveled when 0.72 s has elapsed?

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VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1I looks like an Atwood machine problem, except that there is friction. Have you solved it yet? You can either use laws of motion or energy.

KunoiSlayr
 one year ago
Best ResponseYou've already chosen the best response.0Yes it is an adwood machine scenario. Yes I solved it using some complicated equation [m1*g*sin(theta)m2*gmu*m1*g*cos(theta)]/(m1+m2) but I am still not sure how to draw a free body force diagram.

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1Ok, first m1 will tend to go down, as m1x sin49 > m2 Then forces on m1 are: weight (vertical down) tension (up along the plane) normal reaction (normal to plane) friction (up along the plane since m1 is going down, and equal to µN) Forces on m2 are: weight (vertical down) tension (vertical up, same magnitude as tension acting on m1) Accelerations have different directions but have the same magnitude a. Write down N's 2nd law for m1 and m2 separately, then eliminate tension, then solve for acceleration.
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