Calculate the minimum force required

[ALIAHMAD]

Homework Statement

A block of mass 10kg is placed on a rough horizontal surface with a coefficient of friction of 0.5. the minimum force required to move the block is.(g=10 m/s*s)

Relevant Equations

coefficint of friction * normal reaction = friction force

I tried taking the components but got stuck there on what value of theta I should take

 

[PeroK]

What's $\theta$?

 

[ALIAHMAD]

What's $\theta$?

that's the angle
like sin30 here 30 is theta

 

[PeroK]

that's the angle

What angle?

 

[ALIAHMAD]

What angle?

the angle on which the force is acting on the block with respect to horizontal

 

[PeroK]

the angle on which the force is acting on the block with respect to horizontal

Ah, okay. How far did you get in analysing the forces? Can you post where you got stuck?

 

[ALIAHMAD]

well I drew this diagram first

then for friction to be minimum I took b as my direction force acting on block
then wrote the equations
Fcosθ+N=mg------(1)
Fsinθ-friction of coefficient *N=ma-------(2)

now I don't know what to do after this

 

[PeroK]

well I drew this diagram first View attachment 316634
then for friction to be minimum I took b as my direction force acting on block
then wrote the equations
Fcosθ+N=mg------(1)
Fsinθ-friction of coefficient *N=ma-------(2)

now I don't know what to do after this

You need to put an arrow on forces in your diagram.

Do you think you should pull the block above the horizontal or below the horizontal?

I don't understand the point of A and B.

 

[ALIAHMAD]

You need to put an arrow on forces in your diagram.

Do you think you should pull the block above the horizontal or below the horizontal?

I don't understand the point of A and B.

actually, ignore them
just think that te force applied on the block Is in the direction of B

 

[PeroK]

actually, ignore them
just think that te force applied on the block Is in the direction of B

Are you sure about that? You want to pull the block into the ground? Isn't that going to increase friction and make it harder to move?

 

[ALIAHMAD]

its going to pull the block off the ground.

 

[ALIAHMAD]

hey listen, please
can you just solve it on a piece of paper and post it
it would be very kind of you.

 

[PeroK]

hey listen, please
can you just solve it on a piece of paper and post it
it would be very kind of you.

It would also be cheating! We don't do that here:

https://www.physicsforums.com/help/homeworkhelp/

 

[ALIAHMAD]

It would also be cheating! We don't do that here:

https://www.physicsforums.com/help/homeworkhelp/

oh! okay

 

[ALIAHMAD]

its going to pull the block off the ground.

am i wrong?

 

[Orodruin]

You do not know $\theta$. However, for every value of $\theta$ there will be a corresponding necessary force. The minimum force required will be the smallest of those forces.

 

[ALIAHMAD]

You do not know $\theta$. However, for every value of $\theta$ there will be a corresponding necessary force. The minimum force required will be the smallest of those forces.

yeah i understand it

 

[ALIAHMAD]

sir PeroK
help!

 

[ALIAHMAD]

nailed it man

got the answer

 

[PeroK]

sir PeroK
help!

Your making a fundamental mistake by pulling the block into the ground. That said, if you set up your equations to allow $\theta$ to be positive or negative, then the solution would show that minumim force involves pulling the block at an angle above the horizontal.

My advice is to redraw your diagram with $\theta$ above the horizontal.

The other point is that the minimum force to move the block involves zero accleration. I.e. when the horizontal force on the block equals the frictional force.

 

[ALIAHMAD]

Your making a fundamental mistake by pulling the block into the ground. That said, if you set up your equations to allow $\theta$ to be positive or negative, then the solution would show that minumim force involves pulling the block at an angle above the horizontal.

My advice is to redraw your diagram with $\theta$ above the horizontal.

The other point is that the minimum force to move the block involves zero accleration. I.e. when the horizontal force on the block equals the frictional force.

if i take acceleration to be zero I would give me the force that would not move the block
to make a block move it need to have some acceleration

 

[ALIAHMAD]

Your making a fundamental mistake by pulling the block into the ground. That said, if you set up your equations to allow $\theta$ to be positive or negative, then the solution would show that minumim force involves pulling the block at an angle above the horizontal.

My advice is to redraw your diagram with $\theta$ above the horizontal.

The other point is that the minimum force to move the block involves zero accleration. I.e. when the horizontal force on the block equals the frictional force.

yeah got your point the θ must be above the horizontal.
thank you

 

[PeroK]

if i take acceleration to be zero I would give me the force that would not move the block
to make a block move it need to have some acceleration

With the minimum force the acceleration is negligible. Above that minimum the block will move and below it the block won't move. It's a common idea.

 

[ALIAHMAD]

With the minimum force the acceleration is negligible. Above that minimum the block will move and below it the block won't move. It's a common idea.

considering small acceleration to be negligible we can say that

 

[PeroK]

nailed it man

got the answer
View attachment 316635

I don't see how you got that answer, but it doesn't look right.

 

[PeroK]

Generally, you need calculus to minimise something in a problem with a continuous variable.

 

[ALIAHMAD]

Generally, you need calculus to minimise something in a problem with a continuous variable.

it is a basic level question of my coaching institute's daily practice paper intended for the preparation of JEE
so I think it is ought to be a little easy(not including problems with continuous variable)

 

[ALIAHMAD]

it is a basic level question of my coaching institute's daily practice paper intended for the preparation of JEE
so I think it is ought to be a little easy(not including problems with continuous variable)

you are guiding me the question with understanding of high level physics
and I don't think it applies to these questions

still I am open to learn new things

 

[PeroK]

it is a basic level question of my coaching institute's daily practice paper intended for the preparation of JEE
so I think it is ought to be a little easy(not including problems with continuous variable)

Then, perhaps the answer is just $50 N$? Which is the minumum horizontal force to overcome friction. Considering a force applied at an angle is definitely a more advanced problem.

 

[ALIAHMAD]

this

Then, perhaps the answer is just $25 N$? Which is the minumum horizontal force to overcome friction. Considering a force applied at an angle is definitely a more advanced problem.

this question has option sgiven as follows
a-20√3N
b-50√3N
c-50N
d-20√5 N

 

[PeroK]

this

this question has option sgiven as follows
a-20√3N
b-50√3N
c-50N
d-20√5 N

Okay, the answer is one of those, but not a).

 

[ALIAHMAD]

THE ANSWER GIVEN IS 20√5 maybe by considering that it is close to 25 as it was asked to be minimum

 

[PeroK]

THE ANSWER GIVEN IS 20√5 maybe by considering that it is close to 25 as it was asked to be minimum

You need calculus to get that answer. It's a minimization problem. That generally means calculus is required.

 

[ALIAHMAD]

let me send you the whole question

 

[ALIAHMAD]

the last question (question 19)

 

[ALIAHMAD]

You need calculus to get that answer. It's a minimization problem. That generally means calculus is required.

I understand your point but that's how the question is given
anyways thank you for giving me your precious time

 

[haruspex]

Generally, you need calculus to minimise something in a problem with a continuous variable.

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?

 

[ALIAHMAD]

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?

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)

 

[PeroK]

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)

Or, by deriving that formula.

 

[haruspex]

the combined trig term is sin(θ+𝜶)

Good, so what is the equation you get for F and what is A in terms of $\mu$?

 

[ALIAHMAD]

Good, so what is the equation you get for F and what is A in terms of $\mu$?

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

 

[haruspex]

F=Cos𝜶W/[sin(θ+𝜶)] and A= cos𝜶/μ

I meant, find A purely in terms of $\mu$. $\alpha$ should not appear. But forget that, and instead find $\alpha$ purely in terms of $\mu$.
Use the equations $A=\sin(\alpha), A\mu=\cos(\alpha)$ and eliminate A.
Given your first equation above, what value of $\theta$ minimises F?

 

[Orodruin]

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

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.

Edit: Please note that it is also beneficial for you to write out the entire argument rather than only a result as people will be able to help you better when you go wrong.

 

[PeroK]

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?

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.

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.

 

[haruspex]

That's easy once you know some calculus.

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]

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.

I get that, but a little differentiation goes a long way!

 

[Spector989]

Firstly you should draw a free body diagram . Check what is asked , here you need MINIMUM force so just enough to move the block but not enough to accelerate it . So Force appliead is equal to friction applied . Now how do you know which angle to apply the force on ? Perpendicular? Parallel? Well for now we have no clue so we make the mathematical equations considering it was applied at an angle *theta* and if minimum force was applied Parallel to block the theta would turn out to be zero, the math will do the work for you , all you have to do is arrive at the proper equations . I assume you have been taught some trigonometry as you are solving questions from NLM [A side note :- if you are struggling with step 2 questions i suggest you solve the solved examples then go for step-1 , maybe solve step-3 examples after exercise, they help with concept building :)

 

[Spector989]

You will face some differentiation and integration while solving PYQ , and especially whlie solving advanced questions so it is better to know some basic stuff , when to use it and when to use which (diff or integration) and how to use it.

 

[Orodruin]

So Force appliead is equal to friction applied

This is incorrect.
As has already been discussed in the thread, friction must be equal to the horizontal component of the applied force.

You will face some differentiation and integration while solving PYQ , and especially whlie solving advanced questions so it is better to know some basic stuff , when to use it and when to use which (diff or integration) and how to use it.

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

 

[Spector989]

This is incorrect.
As has already been discussed in the thread, friction must be equal to the horizontal component of the applied force.

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

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 . Sorry