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I'm confused by this question: 1 ...

After resolving forces, I got $T=20g\sin{60} + 0.4\cdot 20g\cos{60} = 209N $

However, this is the answer for part B. After looking through solutions, the minimum tension required is when the frictional force is acting up the slope, lowering the amount of force needed to keep it up.

But, when would the frictional force be acting up the slope? This makes no sense to me. If the force of tension is upwards, surely the friction will always act oppositionally, downwards.

If anything, wouldn't the minimum effect of friction be 0, when the tension is equal to the component of mass parallel to the slope?

Making $T_{\text{min}} = 20g\sin{60}$.

$ F_{\text{friction}} \le \mu F_{\text{normal-reaction}}$

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You have correctly noted that when the frictional force is acting up the slope, it lowers the amount of force needed to keep the package up. So for the part A of the problem you have to consider it with a negative sign. This leads you to a minimum tension of

$$T=20g\sin{60} - 0.4\cdot 20g\cos{60} \approx 131N $$

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