Solving a 100N Weight Block Motion Problem

[goldfish9776]

Homework Statement

The block of weight of 100N is pulled by a rope over a pulley of small block with m kg . A 200N forec also acts horizontally as shown . If thge kinetic coefficient = 0.25 , static coefficient = 0.3 , determine whether the block is moving when the block is in moving or impending motion ?

Homework Equations

The Attempt at a Solution

200cos20 - 0.3( 100x9.81xsin30 + 200sin20 ) =20.2 N , what should i do next ?
how to find the acceleration of 2 blocks , i have no idea , can someone point it out ?

 

[haruspex]

Homework Statement

The block of weight of 100N is pulled by a rope over a pulley of small block with m kg . A 200N forec also acts horizontally as shown . If thge kinetic coefficient = 0.25 , static coefficient = 0.3 , determine whether the block is moving when the block is in moving or impending motion ?

The question statement seems a bit garbled. Please post the exact wording.
It looks like you are only being asked to decide whether it will move. If so, you do not care about the resulting acceleration. Just treat it as a statics problem.
The weight is given in N. It is not a 100kg mass.
Where does the sin 30 come from?
What about the component of gravity down the ramp?

 

[goldfish9776]

The question statement seems a bit garbled. Please post the exact wording.
It looks like you are only being asked to decide whether it will move. If so, you do not care about the resulting acceleration. Just treat it as a statics problem.
The weight is given in N. It is not a 100kg mass.
Where does the sin 30 come from?
What about the component of gravity down the ramp?

given that m = 2kg
sorry , it should be sin20
for the component of force down the slope is 100(9.81)sin20 - 200cos 20 = 733N , what is the next steps?

 

[haruspex]

given that m = 2kg
sorry , it should be sin20
for the component of force down the slope is 100(9.81)sin20 - 200cos 20 = 733N , what is the next steps?

I repeat, it is a 100N weight, not a 100kg mass. Think about that.
In your original expression, you had three forces acting parallel to the slope. Now you are showing only two. How many are there altogether?

 

[goldfish9776]

I repeat, it is a 100N weight, not a 100kg mass. Think about that.
In your original expression, you had three forces acting parallel to the slope. Now you are showing only two. How many are there altogether?

its's stated in the diagram , 100kg mass...

 

[goldfish9776]

so the total force down the plane = 100(9.81)sin20 -200cos20 -0.3 ( 200sin20 +100(9.81)sin20 ) = -149.9N , ?

 

[haruspex]

its's stated in the diagram , 100kg mass...

Ok, so you stated it wrongly in the original post:

The block of weight of 100N

so the total force down the plane = 100(9.81)sin20 -200cos20 -0.3 ( 200sin20 +100(9.81)sin20 ) = -149.9N , ?

Nearly right. Always be suspicious when you see the same force contributing two terms with the same trig function (sine in this case), one with the coefficient of friction and one without.

Also, there is one more force you have not mentioned.

 

[goldfish9776]

Ok, so you stated it wrongly in the original post:
Nearly right. Always be suspicious when you see the same force contributing two terms with the same trig function (sine in this case), one with the coefficient of friction and one without.

Also, there is one more force you have not mentioned.

the total force down the plane = 100(9.81)sin20 -200cos20 -0.3 ( 200sin20 +100(9.81)sin20 )-2(9.81) = -170N ?

 

[goldfish9776]

as Fs= 0.3( 200sin20 +100(9.81)sin20 ) = 297N
resultant force down the plane = 100(9.81)sin20 -200cos20-2(9.81) = 128N , so the object will not moving ,and remain stationary ? as F is less than Fs

 

[haruspex]

the total force down the plane = 100(9.81)sin20 -200cos20 -0.3 ( 200sin20 +100(9.81)sin20 )-2(9.81) = -170N ?

As I hinted in post #7, the second term in 0.3 ( 200sin20 +100(9.81)sin20 ) is wrong. What is the normal force resulting from gravity?
Also, bear in mind that your equation assumes the frictional force is at its maximum value. It could be less.

 

[goldfish9776]

As I hinted in post #7, the second term in 0.3 ( 200sin20 +100(9.81)sin20 ) is wrong. What is the normal force resulting from gravity?
Also, bear in mind that your equation assumes the frictional force is at its maximum value. It could be less.

Fs= 0.3( 200sin20 +100(9.81)cos20 ) = 297N
resultant force down the plane = 100(9.81)sin20 -200cos20-2(9.81) = 128N , so the object will not moving ,and remain stationary ? as F is less than Fs

 

[haruspex]

Fs= 0.3( 200sin20 +100(9.81)cos20 ) = 297N
resultant force down the plane = 100(9.81)sin20 -200cos20-2(9.81) = 128N , so the object will not moving ,and remain stationary ? as F is less than Fs

Yes, though to be more accurate, the Fs there is, as I said, the maximum magnitude of the frictional force. In principle, the block could move either way, so what you need to check is whether |Fs|>|F|, which it is.
The last part of the question is worded strangely. Is this a translation?

when the block is in moving or impending motion

I think it is asking two questions
1. Whether the block will start to move from rest (which you have answered)
2. Whether it would continue to move if it is, by some means, already moving.