Quesiton on Newtons Second Law, Need Help.

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SUMMARY

The discussion centers on calculating the tension in a rope connecting two blocks under the influence of an applied force and friction. Block 1 has a mass of 20 kg, and Block 2 has a mass of 10 kg, with an applied force of 55 N and a total friction force of 44.1 N acting on the system. The correct tension in the rope is determined to be 37 N, which accounts for the acceleration of the blocks rather than simply subtracting the friction force from the applied force. The acceleration of the blocks is calculated to be 0.36 m/s².

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yunuscanemre
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Two blocks of identical material(Block1 20kg, Block2 10kg) are connected by a light rope on a level surface. An applied force of 55Nright causes the blocks to accelerate. While in motion, the magnitude of the force of friction on the block system is 44.1N.


2. Homework Equations
a)Calculate acceleration of the blocks
b)The force of friction on the 10kg block has a magnitude of 14.7 N, calc the tension in the rope connecting the two blocks.




3. The Attempt at a Solution
I found "a" by using:
Fnet=Fapplied-Ffriction
10.9N=30kg x a
a=0.36 m/s^2

For "b"
55N-14.7N=40.3N

Book says the right answer is 37N. I just want to know how its 37N.


I need to understand solving tension problems, I have a test 2morrow.

Thanks :)
 
Last edited:
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Hi yunuscanemre,

yunuscanemre said:
Two blocks of identical material(Block1 20kg, Block2 10kg) are connected by a light rope on a level surface. An applied force of 55Nright causes the blocks to accelerate. While in motion, the magnitude of the force of friction on the block system is 44.1N.


2. Homework Equations
a)Calculate acceleration of the blocks
b)The force of friction on the 10kg block has a magnitude of 14.7 N, calc the tension in the rope connecting the two blocks.




3. The Attempt at a Solution
I found "a" by using:
Fnet=Fapplied-Ffriction
10.9N=30kg x a
a=0.36 m/s^2

For "b"
55N-14.7N=40.3N

Your answer here would be correct if there were no acceleration, because then you would need all of the forces to cancel. However, there is an acceleration here, and so you need to account for that in your equation. Do you see what to do now?
 
thanks I get :))))
 
yunuscanemre said:
thanks I get :))))

Glad to help!
 

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