Calculating m for a Sloping Block with Increased Mass

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In summary: Yes it is correct, but the value of μs is still incorrect. Can you show me the work for finding the mass of the block?Yes it is correct, but the value of μs is still incorrect. Can you show me the work for finding the mass of the block?I think you are doing the problem wrong. You don't need to find the mass of the block, just the maximum value of the mass of the block so that the block does not move. The mass of the block will always be less than this maximum value. I would solve this problem by summing up the forces on the block in the horizontal direction. You should get:##\mu_s M g \
  • #1
goldfish9776
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1.
The block of 50kg with m =30kg is in the verge of moving down the slope where theta = 20degree, determine the new value of m when the mass of block is increased more to 20kg .

Homework Equations

The Attempt at a Solution


I have done in this way , is it okay ?

(50-30)(9.81) = Fs

196.2 = Us (50 x 9.81 cos20)

Us= 0.43

(70-m)(9.81) = 0.43(70 x 9.81 cos20 )

m = 41.7kg
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  • #2
Your way of solving it isn't correct! You cannot add two vertical vectors (weight of M and weight of m) and get a vector with horizontal AND vertical components! In the first equation, you are omitting the angle of the incline, you can't do that.

From Newton's Second Law (assuming acceleration is zero and hanging block is moving down)

∑Fx = 0
⇒ mg - Mgsinθ - Fs - FT + FT = 0
⇒ Fs = g(m - Msinθ)

Since the system is on the verge of moving, the static friction has reached its maximum and thus:

→ Fs(max) = Fs = μsFN

Therefore;

μsFN=g(m - Msinθ)
 
  • #3
faradayscat said:
Your way of solving it isn't correct! You cannot add two vertical vectors (weight of M and weight of m) and get a vector with horizontal AND vertical components! In the first equation, you are omitting the angle of the incline, you can't do that.

From Newton's Second Law (assuming acceleration is zero and hanging block is moving down)

∑Fx = 0
⇒ mg - Mgsinθ - Fs - FT + FT = 0
⇒ Fs = g(m - Msinθ)

Since the system is on the verge of moving, the static friction has reached its maximum and thus:

→ Fs(max) = Fs = μsFN

Therefore;

μsFN=g(m - Msinθ)
using mg-Mg(sin theta) = Fs
i have 30(9.81) -5099.81)(sin20) =Fs
Fs= 126.5 N
126.5=Us(50x9.81xsin20)
Us= 0.75

(70-m)(9.81)= 483.9 , m = 20.7kg , but thge ans given is 41.7 kg ,
is the ans given wrong ?
 
  • #4
goldfish9776 said:
using mg-Mg(sin theta) = Fs
i have 30(9.81) -5099.81)(sin20) =Fs
Fs= 126.5 N
126.5=Us(50x9.81xsin20)
Us= 0.75

(70-m)(9.81)= 483.9 , m = 20.7kg , but thge ans given is 41.7 kg ,
is the ans given wrong ?

My mistake, I forgot to correct your second formula. Like I said, ##(M-m)g## is incorrect, so is ##(70-m)(9.8)##.

This is the equation you should use, as derived previously from Fs=g(m-Msinθ):

→μsFN=g(m - Msinθ)
⇒μsMgcosθ=g(m - Msinθ)

This time M=70kg and m is unknown, you already have μs.

Also, be careful with your trigonometry. The normal force is ##(50)(9.8)cos(20°)## not ##(50)(9.8)sin(20°)##

Do you understand why?
 
Last edited:
  • #5
faradayscat said:
My mistake, I forgot to correct your second formula. Like I said, ##(M-m)g## is incorrect, so is ##(70-m)(9.8)##.

This is the equation you should use, as derived previously from Fs=g(m-Msinθ):

→μsFN=g(m - Msinθ)
⇒μsMgcosθ=g(m - Msinθ)

This time M=70kg and m is unknown, you already have μs.

Also, be careful with your trigonometry. The normal force is ##(50)(9.8)cos(20°)## not ##(50)(9.8)sin(20°)##

Do you understand why?
so , is my method shown above correct?
 
  • #6
No it is incorrect since you forgot to include the angle of the incline into your equations. Can you show me how you got your previous answer? Maybe it will help if I point out some mistakes there.
 
  • #7
faradayscat said:
No it is incorrect since you forgot to include the angle of the incline into your equations. Can you show me how you got your previous answer? Maybe it will help if I point out some mistakes there.
this is what i gt
30(9.81) -5099.81)(sin20) =Fs
Fs= 126.5 N
126.5=Us(50x9.81xsin20)
Us= 0.75

(70-m)(9.81)= 483.9 , m = 20.7kg
 
  • #8
faradayscat said:
No it is incorrect since you forgot to include the angle of the incline into your equations. Can you show me how you got your previous answer? Maybe it will help if I point out some mistakes there.
which part of my qorking that i forgot to include the inclined angle?
 
  • #9
faradayscat said:
No it is incorrect since you forgot to include the angle of the incline into your equations. Can you show me how you got your previous answer? Maybe it will help if I point out some mistakes there.
can you take at look on my another thread , why the tension from 100kg to the roller(up) and the tension from point B to roller doesn't cancel each other?

in the above question , the tension cancel each other?
https://www.physicsforums.com/threads/total-forces-acting-downward-on-the-plane.851538/
 
  • #10
First of all, like I said, normal force is
$$Mgcosθ$$
And NOT
$$Mgsinθ$$

So your answer of ##126.5N## is incorrect as well as your value of the coefficient of static friction of ##0.75##.

Second of all, you should follow my derivation because this equation you are using

##(M-m)g = μMgcosθ##

Is incorrect as well, it should be

##(Msinθ-m)g = μMgcosθ##

Could you please show me the work you did by your own and how you got the correct answer? Please show how you used Newton's Second Law too so I could help you understand better!
 
  • #11
faradayscat said:
First of all, like I said, normal force is
$$Mgcosθ$$
And NOT
$$Mgsinθ$$

So your answer of ##126.5N## is incorrect as well as your value of the coefficient of static friction of ##0.75##.

Second of all, you should follow my derivation because this equation you are using

##(M-m)g = μMgcosθ##

Is incorrect as well, it should be

##(Msinθ-m)g = μMgcosθ##

Could you please show me the work you did by your own and how you got the correct answer? Please show how you used Newton's Second Law too so I could help you understand better!
30(9.81)-50(9.81)sin20 = Fs
Fs=126.5N,
126.5= Us(50x9.81xcos20)
Us= 0.27

(70-m)(9.81) = 0.27 ( 70x9.81xcos20)

m=51.7kg , is it correct?
 
  • #12
Not quite, where are you getting (70-m) from?
 
  • #13
faradayscat said:
Not quite, where are you getting (70-m) from?
when the mass of block is increased more 20kg , caqn you post your working again ?i'm confused now
 
  • #14
faradayscat said:
Not quite, where are you getting (70-m) from?
nvm , focus on the Fs first , you said that my method of finding Fs is incorrect... how to get it correctly ?
 
  • #15
Ok, my bad. I uploaded a picture of my work, hopefully this will help.
 

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  • #16
faradayscat said:
Ok, my bad. I uploaded a picture of my work, hopefully this will help.
that proves my Fs is correct
 
  • #17
##196.2N## is not the value of Fs. It does work to find the mass of ##m##, but you get incorrect values for the coefficient of static friction and frictional force.

Don't forget to use the correct sign for your forces... you should use ##-126.5N## instead of ##126.5N##.
 
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  • #18
goldfish9776 said:
30(9.81)-50(9.81)sin20 = Fs
You have that step right now, but it doesn't match what you did in the 70kg case:
goldfish9776 said:
(70-m)(9.81)
Restructure that expression so that it follows the same form as in the 50kg version. (there are two differences.)
 

1. What is the formula for calculating the slope of a block with increased mass?

The formula for calculating the slope of a block with increased mass is m = F/mg, where m is the slope of the block, F is the force applied, and g is the acceleration due to gravity.

2. How do you determine the increased mass of a block on a sloping surface?

The increased mass of a block on a sloping surface can be determined by measuring the mass of the block on a level surface and then measuring the force required to move the block up the slope. The difference between the two masses is the increased mass.

3. What are the units of measurement for the slope of a block with increased mass?

The slope of a block with increased mass is typically measured in kilograms per meter (kg/m).

4. How does the slope of a block with increased mass affect its stability?

The slope of a block with increased mass can affect its stability by increasing the force required to move the block, making it more difficult to maintain its position on the slope. It can also change the center of mass of the block, potentially making it more prone to tipping over.

5. What are some factors that can affect the slope of a block with increased mass?

Some factors that can affect the slope of a block with increased mass include the weight and distribution of the added mass, the angle of the slope, and the coefficient of friction between the block and the slope.

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