I cannot draw the F.B.D for you here, but assume that m1 is the mass being pulled by force F, and that m2 is the mass that is trailing behind, pulled by rope tension T. T also ags as a reverse force on mass m1.
The equations of motion are
F - Ff1 - T = m1*a
T - Ff2 = m2*a
Adding the two equations together cancels out the t and lets you solve for a.
F - Ff1 - Ff2 = (m1 + m2)*a
a = (F - Ff1 - Ff2)/(m1 + m2)
You already have written the correct equation for the Ffk friction terms.
Once you a from the above equation, you can solve for T.
T = Ff2 + m2*(F - Ff1 - Ff2)/(m1 + m2)
2 blocks are connected by a rope of negligible mass are being dragged by a horizontal force. Suppose F= 68.0N, m1= 12.0kg, m2= 18.0kg and the coefficient of kinetic friction between each block and the surface is 0.100.
a.) draw free body diagram for each block
b.) determine the tension T and the magnitude of the acceleration of the system.
Ffk= uk * Fn (Ffk=force of kinetic friction) (uk= coefficient of friction) (Fn= normal force)
F-T= m2*a
T-f= m1a or is it T= m1*a (I changed the equation that I saw for a system with objects being pulled but they said it was T= m1*a however I'm thinking that this doesn't include friction into it since the one with the first box connected directly to the rope being pulled does take into account the tension force and it subtracts it from the force that pulls the box ...so I was thinking maybe they didn't include the friction since it would also be a opposing force to the tension..am I correct?)
a= F/m1 + m2
I really really really really need help in how i incorperate the coefficient of friction into this problem..I was thinking that that that would be the force it is pulled with but I really don't know what to do..sinc they give you a force that the boxes are pulled with
Can someone give me a push in the right direction...especially how to relate the forces to the coefficient of friction..
THANKS
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THANKS SO MUCH DrWls !! =D
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