Quesiton on Newtons Second Law, Need Help.

AI Thread Summary
Two blocks of different masses (20kg and 10kg) are connected and subjected to a 55N force, with a friction force of 44.1N acting on the system. The net force was calculated as 10.9N, leading to an acceleration of 0.36 m/s². To find the tension in the rope, the friction on the 10kg block (14.7N) must be considered alongside the acceleration, which was initially overlooked. The correct tension in the rope is 37N, as confirmed by the book. Understanding the role of acceleration is crucial for solving tension problems in this context.
yunuscanemre
Messages
2
Reaction score
0
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:
Physics news on Phys.org
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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top