Tension between two sliding blocks

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SUMMARY

The discussion focuses on calculating the tension in a cord connecting two blocks sliding down an inclined plane at an angle of 28 degrees. Given the masses of both blocks as 5.5 kg and the coefficients of friction as μA = 0.19 and μB = 0.30, the calculated tension is approximately 2.722 N. The solution is deemed correct in principle, but the acceleration was rounded excessively, affecting the precision of the tension value. A more accurate approach involves directly canceling acceleration to find tension.

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Homework Statement



The problem:

gUEoZ.jpg


Two blocks made of different materials connected together by a thin cord, slide down a plane ramp inclined at an angle θ to the horizontal as shown in the figure (block B is above block A). The masses of the blocks are mA and mB and the coefficients of friction are μA and μB.

If mA=mB=5.5kg, and μA = 0.19 and μB = 0.30, determine the tension in the cord, for an angle ##θ = 28^{\circ}##

Homework Equations



Newton's Laws

The Attempt at a Solution



My Attempt:After setting up free body diagrams (not shown), I found the acceleration ##a\approx2.5~\frac{m}{s^2}## along with the following equation for T:

$$sin(\theta )mg-T-\mu _acos(\theta)mg=ma$$

$$\Rightarrow T=m(sin(\theta)g-\mu _acos(\theta)g-a)\Rightarrow$$

##T\approx 2.722 ~\textrm{N}## in magnitude

Is this the correct answer?
 
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END said:

Homework Statement



The problem:

gUEoZ.jpg


Two blocks made of different materials connected together by a thin cord, slide down a plane ramp inclined at an angle θ to the horizontal as shown in the figure (block B is above block A). The masses of the blocks are mA and mB and the coefficients of friction are μA and μB.

If mA=mB=5.5kg, and μA = 0.19 and μB = 0.30, determine the tension in the cord, for an angle ##θ = 28^{\circ}##

Homework Equations



Newton's Laws

The Attempt at a Solution



My Attempt:


After setting up free body diagrams (not shown), I found the acceleration ##a\approx2.5~\frac{m}{s^2}## along with the following equation for T:

$$sin(\theta )mg-T-\mu _acos(\theta)mg=ma$$

$$\Rightarrow T=m(sin(\theta)g-\mu _acos(\theta)g-a)\Rightarrow$$

##T\approx 2.722 ~\textrm{N}## in magnitude

Is this the correct answer?

The solution is correct in principle, but you rounded off the acceleration too much, therefore the second digit of T is inaccurate. You could have canceled a and get T directly.

ehild
 

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