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I understand your point but that's how the question is givenPeroK said:You need calculus to get that answer. It's a minimization problem. That generally means calculus is required.

anyways thank you for giving me your precious time

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In summary: misconception... to think that the block will always move when there is a force that is greater than the friction.

- #36

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I understand your point but that's how the question is givenPeroK said:You need calculus to get that answer. It's a minimization problem. That generally means calculus is required.

anyways thank you for giving me your precious time

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But not always, and not here.PeroK said:Generally, you need calculus to minimise something in a problem with a continuous variable.

@ALIAHMAD , if you write the two force balance equations and eliminate the normal force you should end up with ##F(\cos(\theta)+\mu\sin(\theta))=\mu W##.

The trick then is to multiply by a constant such that the trig terms can be collapsed into one: ##F(A\cos(\theta)+A\mu\sin(\theta))=A\mu W## where ##A=\sin(\alpha), A\mu=\cos(\alpha)##. Can you see how to choose A such that ##\alpha## exists? Can you then collapse that into a single trig term, and then spot what value of ##\theta## minimises F?

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yeah I can do theseharuspex said:But not always, and not here.

@ALIAHMAD , if you write the two force balance equations and eliminate the normal force you should end up with ##F(\cos(\theta)+\mu\sin(\theta))=\mu W##.

The trick then is to multiply by a constant such that the trig terms can be collapsed into one: ##F(A\cos(\theta)+A\mu\sin(\theta))=A\mu W## where ##A=\sin(\alpha), A\mu=\cos(\alpha)##. Can you see how to choose A such that ##\alpha## exists? Can you then collapse that into a single trig term, and then spot what value of ##\theta## minimises F?

like the combined trig term is sin(θ+𝜶)

my teacher told me that considering θ will make the question more complicated as for every angle there is some minimum and then to find the minimalist of them is very hard

he told me that these types of questions are solved by using this formula

F(min)=mgμ/√(1+μ^2)

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Or, by deriving that formula.ALIAHMAD said:yeah I can do these

like the combined trig term is sin(θ+𝜶)

my teacher told me that considering θ will make the question more complicated as for every angle there is some minimum and then to find the minimalist of them is very hard

he told me that these types of questions are solved by using this formula

F(min)=mgμ/√(1+μ^2)

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Good, so what is the equation you get for F and what is A in terms of ##\mu##?ALIAHMAD said:the combined trig term is sin(θ+𝜶)

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F=Cos𝜶W/[sin(θ+𝜶)] and A= cos𝜶/μharuspex said:Good, so what is the equation you get for F and what is A in terms of ##\mu##?

just wanted to know how that formula is derived as I don't see any clue on it

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I meant, find A purely in terms of ##\mu##. ##\alpha## should not appear. But forget that, and instead find ##\alpha## purely in terms of ##\mu##.ALIAHMAD said:F=Cos𝜶W/[sin(θ+𝜶)] and A= cos𝜶/μ

Use the equations ##A=\sin(\alpha), A\mu=\cos(\alpha)## and eliminate A.

Given your first equation above, what value of ##\theta## minimises F?

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It seems you have not really gotten the point here. The point is to introduce a constant ##A##, expressed only in the variable ##\mu##, that allows you to rewrite your expression using a single trigonometric function with ##\theta+\alpha## as its argument.ALIAHMAD said:F=Cos𝜶W/[sin(θ+𝜶)] and A= cos𝜶/μ

just wanted to know how that formula is derived as I don't see any clue on it

Edit: Please note that it is also beneficial

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Here's a simpler alternative. To minimise ##F## we need to maximise ##\cos(\theta)+\mu\sin(\theta)##. That's easy once you know some calculus.haruspex said:But not always, and not here.

@ALIAHMAD , if you write the two force balance equations and eliminate the normal force you should end up with ##F(\cos(\theta)+\mu\sin(\theta))=\mu W##.

The trick then is to multiply by a constant such that the trig terms can be collapsed into one: ##F(A\cos(\theta)+A\mu\sin(\theta))=A\mu W## where ##A=\sin(\alpha), A\mu=\cos(\alpha)##. Can you see how to choose A such that ##\alpha## exists? Can you then collapse that into a single trig term, and then spot what value of ##\theta## minimises F?

Failing that, we let ##\mu = \tan \alpha## (for some ##0 < \alpha < \frac \pi 2)## and aim to maximise ##\cos(\theta)+\tan(\alpha) \sin(\theta)##. One more tweak and a trig identity should do it.

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Sure, but the point of my post was, in view of post #28, to show it can be done without any calculus. Besides, it is a useful trick in another context: solving equations involving a mix of trig functions of the same unknown.PeroK said:That's easy once you know some calculus.

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I get that, but a little differentiation goes a long way!haruspex said:Sure, but the point of my post was, in view of post #28, to show it can be done without any calculus. Besides, it is a useful trick in another context: solving equations involving a mix of trig functions of the same unknown.

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This is incorrect.Spector989 said:So Force appliead is equal to friction applied

As has already been discussed in the thread, friction must be equal to the

That’s fine but — again as already discussed in the thread — differentiation is not necessary in this problem.Spector989 said:

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Mb i should have mentioned it is equal to the horizontal component and Normal is equal to mg - minus vertical component (given force is applied above horizontal) and i simply siad that so that he wouldn't back off from differentiation and integration too much , in jee 11th you will face some calculus problems while solving some physics questions. But again it is my fault because i was not clear . SorryOrodruin said:This is incorrect.

As has already been discussed in the thread, friction must be equal to thehorizontal componentof the applied force.

That’s fine but — again as already discussed in the thread — differentiation is not necessary in this problem.

- #51

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No, when the block moves with a constant velocity it has zero acceleration.ALIAHMAD said:if i take acceleration to be zero I would give me the force that would not move the block

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Then it would be kinetic friction that is relevant. The question relates to overcoming static friction.Mister T said:No, when the block moves with a constant velocity it has zero acceleration.

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