Static Friction/Determining Force

In summary, the problem involves a 2 kg block at rest on a horizontal surface with a coefficient of static friction of 0.3. The minimum force F that must be applied to the block at an angle of 53 degrees below horizontal to move the block needs to be determined. The equations involved are the vertical force balance, horizontal force balance, and the friction relationship at the verge of block slipping. These equations are used to find the force F.
  • #1
Masrat_A

Homework Statement


A 2 kg block is at rest on a horizontal surface with which its coefficient of static friction is 0.3. Determine the minimum force F that must be applied to the block at an angle of 53 degrees below horizontal in order to move the block (see Figure 2).

Figure: http://i.imgur.com/OeiNvjZh.jpg

Homework Equations


Please see below.

The Attempt at a Solution


I was unable to take good notes this past lecture because I was not at a decent seat, and based on what I do have, this is what I had come up with:

##(0.3)(2)(9.8)cos53^o##
##= (5.88)(0.60)##
##= 3.5 N##

I know this cannot be right.

Could someone please explain to me how we could find the minimum force (while pointing towards a specific formula I should look into)?
 
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  • #2
Which forces are acting on the block?
Draw a free-body diagram of the block showing all those forces.
 
  • #3
I'm sorry, but could you please rephrase? There is a diagram in the image I linked.
 
  • #4
Figure 2 only shows the applied force, you need to draw a free body diagram showing all the forces.
 
  • #5
I know there is gravity, but are there any other forces I could be missing?
 
  • #6
Figure 2 doesn't show the friction force either.

Aside: I have a thread running in the moderators section entitled "Reluctance to make drawings". Can you guess what it's about :-)
 
  • #7
Does this seem correct? http://i.imgur.com/mtJcoAeh.jpg

I'm assuming the friction is ##30N##, as the coefficient is stated to be 0.3 (am I wrong?). As for gravity, I've multiplied ##m## with ##g##, resulting in ##20N##.

What would be my next step?

CWatters said:
Aside: I have a thread running in the moderators section entitled "Reluctance to make drawings". Can you guess what it's about :-)

I apologize, haha.
 
  • #8
You're still missing a force.
 
  • #9
Chestermiller said:
You're still missing a force.

As the block is on a surface, would this be normal force, which is pushing the block towards upward direction?

##F_{frict} = 0.3 * F_{norm}##
##30 = 0.3 * F_{norm}##
##100N = F_{norm}##
 
  • #10
Masrat_A said:
As the block is on a surface, would this be normal force, which is pushing the block towards upward direction?

##F_{frict} = 0.3 * F_{norm}##
##30 = 0.3 * F_{norm}##
##100N = F_{norm}##
Yes, you had omitted the normal force. So, if F is the pushing force, what are its horizontal and vertical components? In terms of F (algebraically), what are your force balances in the horizontal and vertical directions (calling f the frictional force)?
 
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  • #11
+1

In this case the normal force isn't just the reaction force due to gravity.
 
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  • #12
May I please have an idea as to how I could go about tackling the horizontal and vertical components?
 
  • #13
Masrat_A said:
May I please have an idea as to how I could go about tackling the horizontal and vertical components?
horizontal component = ##F\cos{53}##

vertical component = ##F\sin{53}##

You've had trigonometry, right? You form a right triangle with the force F as the hypotenuse.
 
  • #14
Chestermiller said:
horizontal component = ##F\cos{53}##

vertical component = ##F\sin{53}##

You've had trigonometry, right? You form a right triangle with the force F as the hypotenuse.

I don't have much experience with trigonometry, sadly.

If pushing force is hypothenuse, would normal force and friction be adjacent and opposite respectively? After I do find hypothenuse and calculate the vertical and horizontal components, which are listed below, where shall I then go to find minimum force?

##100^2 + 30^2 = 10,900 ##
##F = 104.4##

##104.4\sin{53} = 83.38##
##104.4\cos{53} = 62.83##
 
  • #15
Masrat_A said:
I don't have much experience with trigonometry, sadly.

If pushing force is hypothenuse, would normal force and friction be adjacent and opposite respectively? After I do find hypothenuse and calculate the vertical and horizontal components, which are listed below, where shall I then go to find minimum force?

##100^2 + 30^2 = 10,900 ##
##F = 104.4##

##104.4\sin{53} = 83.38##
##104.4\cos{53} = 62.83##
Where did the 30 N come from? Why is it even in there?

The vertical component of the applied force is opposite, and the horizontal component of the applied force is adjacent. The frictional force isn't even part of resolving the applied force into components.
 
  • #16
Chestermiller said:
Where did the 30 N come from? Why is it even in there?

The vertical component of the applied force is opposite, and the horizontal component of the applied force is adjacent. The frictional force isn't even part of resolving the applied force into components.

Oh, I'm sorry. ##30N## is the frictional force.

Thank you for clearing that! If we don't already know what F equates to, how would we be able to calculate ##Fcos{53}## and ##Fsin{53}## and figure out the vertical and horizontal components?
 
  • #17
Masrat_A said:
Oh, I'm sorry. ##30N## is the frictional force.

Thank you for clearing that! If we don't already know what F equates to, how would we be able to calculate ##Fcos{53}## and ##Fsin{53}## and figure out the vertical and horizontal components?
Temporarily, we are leaving it algebraic. We will solve for F next, after we set up our force balance equations.
 
  • #18
Do any of these seem accurate?

##F_{norm} = F_{grav} + F_y##
##F_{norm} - F_{grav} = F_y##
##100 - 20 = F_y##
##F_y = 80##

##F_y = F\sin{53}##
##F_y/\sin{53} = F##
##80/0.80 = F##
##F = 100##

##F_x - F_{frict} = F_{net}##
##F_x = F_{net} + F_{frict}##
##F_x = 250 + 30##
##F_x = 280##

##100\cos{53} = 60.18##
##100\sin{53} = 79.86##
 
  • #19
Masrat_A said:
Do any of these seem accurate?

##F_{norm} = F_{grav} + F_y##
##F_{norm} - F_{grav} = F_y##
##100 - 20 = F_y##
##F_y = 80##

##F_y = F\sin{53}##
##F_y/\sin{53} = F##
##80/0.80 = F##
##F = 100##

##F_x - F_{frict} = F_{net}##
##F_x = F_{net} + F_{frict}##
##F_x = 250 + 30##
##F_x = 280##

##100\cos{53} = 60.18##
##100\sin{53} = 79.86##
Actually, none of this makes sense to me. Here's what does make sense to me:

VERTICAL FORCE BALANCE: $$F\sin{53}+20-N=0$$where N is the upward normal force exerted by the surface on the block.

HORIZONTAL FORCE BALANCE: $$F\cos{53}-F_{friction}=0$$

FRICTION RELATIONSHIP (AT VERGE OF BLOCK SLIPPING): $$F_{friction}=\mu N$$where ##\mu## is the coefficient of static friction.

Do these equations make sense to you? Do you think you can combine these equations to solve for the force F?
 
  • #20
Chestermiller said:
Actually, none of this makes sense to me. Here's what does make sense to me:

VERTICAL FORCE BALANCE: $$F\sin{53}+20-N=0$$where N is the upward normal force exerted by the surface on the block.

HORIZONTAL FORCE BALANCE: $$F\cos{53}-F_{friction}=0$$

FRICTION RELATIONSHIP (AT VERGE OF BLOCK SLIPPING): $$F_{friction}=\mu N$$where ##\mu## is the coefficient of static friction.

Do these equations make sense to you? Do you think you can combine these equations to solve for the force F?

They definitely do! Thank you very much! Would you please mind checking if I had applied them correctly?

##F\sin{53}+20-N=0##
##F\sin{53}+20=N##
##0.8F+20=N##

##F_{friction}=\mu N##
##F_{friction}=0.3(0.8F+20)##
##F_{friction}=0.24F+6##

##F\cos{53}-F_{friction}=0##
##0.6F=F_{friction}##
##0.6F=0.24F+6##
##0.36F=6##
##F=16.7##
 
  • #21
Masrat_A said:
They definitely do! Thank you very much! Would you please mind checking if I had applied them correctly?

##F\sin{53}+20-N=0##
##F\sin{53}+20=N##
##0.8F+20=N##

##F_{friction}=\mu N##
##F_{friction}=0.3(0.8F+20)##
##F_{friction}=0.24F+6##

##F\cos{53}-F_{friction}=0##
##0.6F=F_{friction}##
##0.6F=0.24F+6##
##0.36F=6##
##F=16.7##
Confirmed.
 

1. What is static friction?

Static friction is the force that prevents two surfaces from sliding past each other when one surface is trying to move relative to the other.

2. How is static friction different from kinetic friction?

Static friction only occurs when there is no relative motion between the two surfaces, while kinetic friction occurs when there is relative motion between the two surfaces.

3. How is static friction measured?

Static friction can be measured by applying a force to one surface and measuring the force needed to overcome the static friction and cause the two surfaces to start moving relative to each other.

4. What factors affect the magnitude of static friction?

The magnitude of static friction is affected by the nature of the two surfaces in contact, the amount of force being applied, and the roughness of the surfaces.

5. How is the coefficient of static friction determined?

The coefficient of static friction, which represents the ratio of the maximum static friction force to the normal force between the two surfaces, can be determined experimentally by measuring the force required to overcome static friction at various angles and using the equation μs = F/max.

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