Static Friction/Determining Force

[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)?

 

[cnh1995]

Which forces are acting on the block?
Draw a free-body diagram of the block showing all those forces.

 

[Masrat_A]

I'm sorry, but could you please rephrase? There is a diagram in the image I linked.

 

[CWatters]

Figure 2 only shows the applied force, you need to draw a free body diagram showing all the forces.

 

[Masrat_A]

I know there is gravity, but are there any other forces I could be missing?

 

[CWatters]

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 :-)

 

[Masrat_A]

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?

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.

 

[Chestermiller]

You're still missing a force.

 

[Masrat_A]

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?

$F_{frict} = 0.3 * F_{norm}$
$30 = 0.3 * F_{norm}$
$100N = F_{norm}$

 

[Chestermiller]

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)?

 

[CWatters]

+1

In this case the normal force isn't just the reaction force due to gravity.

 

[Masrat_A]

May I please have an idea as to how I could go about tackling the horizontal and vertical components?

 

[Chestermiller]

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.

 

[Masrat_A]

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?

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

$104.4\sin{53} = 83.38$
$104.4\cos{53} = 62.83$

 

[Chestermiller]

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.

 

[Masrat_A]

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?

 

[Chestermiller]

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.

 

[Masrat_A]

Do any of these seem accurate?

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

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

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

$100\cos{53} = 60.18$
$100\sin{53} = 79.86$

 

[Chestermiller]

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?

 

[Masrat_A]

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?

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

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

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

 

[Chestermiller]

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.