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Three blocks, with masses m1 = 5.90 kg, m2 = 5.50 kg, and m3 = 7.10 kg, are pull on a horizontal frictionless surface by a 18.00 N force that makes a 39 o angle (θ) with the horizontal (see figure). What is the magnitude of the tension between the m2 and m3 blocks?

https://general.physics.rutgers.edu/gifs/CJ/4-44.jpg

Hello Neal,

The applied force is not horizontal, so we need to calculate what component of the 18 N is in the horizontal direction.

F_horiz = 18.00 N*cos39 = 13.99 N

The sum of the 3 masses to be accelerated is 18.5 kg. They will accelerate according to Newton's 2nd Law:

F = m*a so in this case

a = F_horiz/m = 13.99 N/18.5 kg = 0.7562 N/kg = 0.7562 m/s^2

So what is the tension in the cord pulling m3? It is enough to accelerate a 7.10 kg mass at a rate of 0.7562 m/s^2.

You can do the rest of this! Use Newton's 2nd again, where this time F is the tension, T, and m is the mass m3.

I hope this helps,

Steve

Physics

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