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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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Imagine the man is lowering the package. Now the package moves down, and the force of friction points up. As the man starts to pull a little bit harder on the string, the package stops - but the force of friction is still supporting the package (preventing it from sliding down).

Perhaps an easier analogy would be this: if there was a hook with a breaking strength of 50 N and the package was hanging from that hook, how much less hard can the man pull on the string without the package sliding down (and the hook breaking)? Think of friction as a "hook with a breaking strength". If the object is not moving the force of friction will be opposing the direction "in which you are almost moving".

Does that help?

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  • $\begingroup$ Thanks, this did help me conceptualise the situation. The box is just about to slip down. $\endgroup$ – Tobi Feb 8 '17 at 21:55
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Friction opposes relative motion.

The box trying to move up or down the slope will be opposing friction. the direction of friction will change depending on what is happening.

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