Friction -- Newton's laws of motion

[Vv anand]

Homework Statement

A student is cleaning a block board by moving a light duster up and down as shown. (Coefficient of friction between the duster and board is $MU$
The duster will not move no matter how large the force is if ?

Homework Equations

No equations given

The Attempt at a Solution

Attempted my making free body diagrams but am unable to solve it

 

[Vv anand]

Homework Statement

A student is cleaning a block board by moving a light duster up and down as shown. (Coefficient of friction between the duster and board is $mu$
The duster will not move no matter how large the force is if ?

Homework Equations

No equations given

The Attempt at a Solution

Attempted my making free body diagrams but am unable to solve it

 

[Nidum]

Hint : The duster is seeing two components of the applied force - one pressing it against the blackboard and one trying to slide it up ?.

Best to follow the usual free body and forces method of solving this time but just for interest this problem can be answered by inspection if you can spot how to use a simple formula from friction force theory . We can explore this later if you wish .

 

[Vv anand]

View attachment 206320Hint : The duster is seeing two components of the applied force - one pressing it against the blackboard and one trying to slide it up ?.

Best to follow the usual free body and forces method of solving this time but just for interest this problem can be answered by inspection if you can spot how to use a simple formula from friction force theory . We can explore this later if you wish .

Will i consider mg in the fbd?
Edit:Just because the duster is light will the ng of duster be considered

 

[scottdave]

Will i consider mg in the fbd?
Edit:Just because the duster is light will the m*g of duster be considered

I would guess that they are telling you that it is light, meaning that the weight (m*g) of the duster is very small compared to the other forces. Remember that the tan() function is opposite / adjacent. Remember what mu represents: (friction force) / (normal force) {edited}, but since this is static friction (not moving), the friction force will be just enough to keep it from moving, and is parallel to the surface in the opposite direction of parallel component of applied force)

It looks like a multiple choice question. Do any of the choices include m*g ?

 

[Vv anand]

No thats

I would guess that they are telling you that it is light, meaning that the weight (m*g) of the duster is very small compared to the other forces. Remember that the tan() function is opposite / adjacent. Remember what mu represents: (friction force) / (normal force) {edited}, but since this is static friction (not moving), the friction force will be just enough to keep it from moving, and is parallel to the surface in the opposite direction of parallel component of applied force)

It looks like a multiple choice question. Do any of the choices include m*g ?

that's why is asked

 

[Nidum]

The question specifies a light duster so we can reasonably interpret that as saying it has zero mass .

The duster has an applied force acting on it which can be split into two components .

What is the magnitude and direction of the friction force tending to or actually preventing movement of the duster ?

Can you now draw a free body diagram showing how all three forces act on the duster ?

Hint : What is the relationship between normal force and friction force ?

 

[Jay Kumar]

If we make components of F, we'll get F cos(beta) on the +ve y-axis and F sin(beta) in the +ve x-axis...
Since the duster is light we won't consider the mass of the duster!
Since the duster has to move in the +ve y direction for it to clean the board, a frictional force will also act in the -ve y-axis direction trying to oppose it...
Now as the sin component of force F is acting on the black board! A normal reaction "N" will be present normal to the duster and the black board, equal to F sin(beta)
As friction f = (MU)N and N = F sin(beta) ,
f = (MU)F sin(beta)
Now let us think the opposite way (with reference to the question), if we need to move the duster... The net vertical force should be GREATER than 0... Right?
So if we write,
F cos(beta) - f > 0
Then,
F cos(beta) > (MU)F sin(beta)
Once we cancel the forces,
Tan(beta) < 1/(MU)
So this is the condition when the duster WILL MOVE!
So for the duster to NOT move at all,
Tan(beta) > 1/(MU)
I hope this helps...! :)