Tension between two sliding blocks

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The problem involves two blocks sliding down an inclined plane, connected by a cord, with specified masses and coefficients of friction. The calculated acceleration is approximately 2.5 m/s², leading to a tension value of about 2.722 N. However, the accuracy of the tension calculation is questioned due to rounding in the acceleration value. It is suggested that a more precise approach could yield a more accurate result for tension. The discussion emphasizes the importance of precision in calculations involving Newton's laws.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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