## Machines B/C

oliviasl
Member
Posts: 1
Joined: December 19th, 2020, 10:11 am
Division: C
State: CA
Pronouns: She/Her/Hers
Has thanked: 0
Been thanked: 0

### Re: Machines B/C

Ttonyxx wrote:
February 9th, 2021, 11:14 am
Okay, I'll guess I'll go since no one else is going.

1) Let's start easy: What are the three Archimedean simple machines?
2) A little harder: What is the ideal mechanical advantage of a 0.112in-40 UNC-3A x 0.5 screw? Explain what each of the numbers in the screw callout means.
3) In the diagram shown below, the block has a mass m = 100kg and there is a coefficient of friction μ = 0.203 between the surfaces on the top and bottom of the wedge. The angle of the wedge (between the two contact surfaces) is 15°. Assume every other surface is frictionless and the only moveable pieces are the wedge and the block. Calculate the force P required to raise the block.
1) Lever, screw, pulley
2) 0.112in refers to the major diameter. 40 means there are 40 threads per inch. UNC means the screw is a Unified National Coarse Thread screw. The 3 indicates the thread class, with 3 being the class with the lowest tolerance. "A "refers to external threads. The 0.5 means that the screw is half an inch long.
3) I got P = 687.7 N
Last edited by pikachu4919 on February 13th, 2021, 7:24 am, edited 1 time in total.

Ttonyxx
Member
Posts: 9
Joined: February 27th, 2019, 5:32 pm
Division: C
State: CA
Pronouns: He/Him/His
Has thanked: 1 time
Been thanked: 0
Contact:

### Re: Machines B/C

oliviasl wrote:
February 12th, 2021, 4:57 pm
Ttonyxx wrote:
February 9th, 2021, 11:14 am
Okay, I'll guess I'll go since no one else is going.

1) Let's start easy: What are the three Archimedean simple machines?
2) A little harder: What is the ideal mechanical advantage of a 0.112in-40 UNC-3A x 0.5 screw? Explain what each of the numbers in the screw callout means.
3) In the diagram shown below, the block has a mass m = 100kg and there is a coefficient of friction μ = 0.203 between the surfaces on the top and bottom of the wedge. The angle of the wedge (between the two contact surfaces) is 15°. Assume every other surface is frictionless and the only moveable pieces are the wedge and the block. Calculate the force P required to raise the block.
1) Lever, screw, pulley
2) 0.112in refers to the major diameter. 40 means there are 40 threads per inch. UNC means the screw is a Unified National Coarse Thread screw. The 3 indicates the thread class, with 3 being the class with the lowest tolerance. "A "refers to external threads. The 0.5 means that the screw is half an inch long.
3) I got P = 687.7 N
Yup! Everything is correct, your turn.

RiverWalker88
Exalted Member
Posts: 103
Joined: February 24th, 2020, 7:14 pm
Division: C
State: NM
Pronouns: He/Him/His
Has thanked: 79 times
Been thanked: 143 times
Contact:

### Re: Machines B/C

Revive?

Here's a problem I'd recommend for everyone that has some basic knowledge of trig (i.e. how the trig functions relate to a triangle) doing machines, as it really helps solidify and conceptualize the formula for IMA of a wedge (and you get the warm fuzzy feeling of having come up with something you can actually use during the event).

For both of the following wedges, the Ideal Mechanical Advantage can be described as $IMA = \frac{E}{R}$. Using this, determine a formula for the IMA of both of these wedges in terms of the angle of the wedge $\theta$. Wedge (a) is a right triangle, and wedge (b) is isosceles. Show and/or explain how you got your formula.

Machines QM Wedges.jpg (17.13 KiB) Viewed 106 times
Socorro High School (2021 Events: Astro, Chem Lab, Circuit Lab, Codybusters, Detector, ExDes, Machines)
2021 Socorro High Invitational Director (Thanks to those who competed and volunteered, it was fun!)
RiverWalker88's Userpage (Mostly Complete)

Lemonism Forever

Ttonyxx
Member
Posts: 9
Joined: February 27th, 2019, 5:32 pm
Division: C
State: CA
Pronouns: He/Him/His
Has thanked: 1 time
Been thanked: 0
Contact:

### Re: Machines B/C

Revive?

Here's a problem I'd recommend for everyone that has some basic knowledge of trig (i.e. how the trig functions relate to a triangle) doing machines, as it really helps solidify and conceptualize the formula for IMA of a wedge (and you get the warm fuzzy feeling of having come up with something you can actually use during the event).

For both of the following wedges, the Ideal Mechanical Advantage can be described as $IMA = \frac{E}{R}$. Using this, determine a formula for the IMA of both of these wedges in terms of the angle of the wedge $\theta$. Wedge (a) is a right triangle, and wedge (b) is isosceles. Show and/or explain how you got your formula.

Machines QM Wedges.jpg
I agree, this becomes pretty useful in competition. Here's my solution:
$\\IMA=\dfrac{E}{R}\\tan(\frac{\theta}{2})=\frac{R/2}{E}\\2tan(\frac{\theta}{2})=\frac{R}{E}\\\frac{1}{2tan(\frac{\theta}{2})}=\frac{E}{R}\\IMA=\frac{1}{2tan(\frac{\theta}{2})}$

RiverWalker88
Exalted Member
Posts: 103
Joined: February 24th, 2020, 7:14 pm
Division: C
State: NM
Pronouns: He/Him/His
Has thanked: 79 times
Been thanked: 143 times
Contact:

### Re: Machines B/C

Ttonyxx wrote:
February 25th, 2021, 9:12 pm
Revive?

Here's a problem I'd recommend for everyone that has some basic knowledge of trig (i.e. how the trig functions relate to a triangle) doing machines, as it really helps solidify and conceptualize the formula for IMA of a wedge (and you get the warm fuzzy feeling of having come up with something you can actually use during the event).

For both of the following wedges, the Ideal Mechanical Advantage can be described as $IMA = \frac{E}{R}$. Using this, determine a formula for the IMA of both of these wedges in terms of the angle of the wedge $\theta$. Wedge (a) is a right triangle, and wedge (b) is isosceles. Show and/or explain how you got your formula.

Machines QM Wedges.jpg
I agree, this becomes pretty useful in competition. Here's my solution:
$\\IMA=\dfrac{E}{R}\\tan(\frac{\theta}{2})=\frac{R/2}{E}\\2tan(\frac{\theta}{2})=\frac{R}{E}\\\frac{1}{2tan(\frac{\theta}{2})}=\frac{E}{R}\\IMA=\frac{1}{2tan(\frac{\theta}{2})}$
Looks great for wedge B! Wedge A is a little different (not sure if you just missed it or...?)

And yeah, I did the derivation a couple of times and the formula is completely solidified in my head–again would recommend trying this out to anyone who does Machines.
Socorro High School (2021 Events: Astro, Chem Lab, Circuit Lab, Codybusters, Detector, ExDes, Machines)
2021 Socorro High Invitational Director (Thanks to those who competed and volunteered, it was fun!)
RiverWalker88's Userpage (Mostly Complete)

Lemonism Forever

Ttonyxx
Member
Posts: 9
Joined: February 27th, 2019, 5:32 pm
Division: C
State: CA
Pronouns: He/Him/His
Has thanked: 1 time
Been thanked: 0
Contact:

### Re: Machines B/C

RiverWalker88 wrote:
February 25th, 2021, 9:34 pm
Ttonyxx wrote:
February 25th, 2021, 9:12 pm
Revive?

Here's a problem I'd recommend for everyone that has some basic knowledge of trig (i.e. how the trig functions relate to a triangle) doing machines, as it really helps solidify and conceptualize the formula for IMA of a wedge (and you get the warm fuzzy feeling of having come up with something you can actually use during the event).

For both of the following wedges, the Ideal Mechanical Advantage can be described as $IMA = \frac{E}{R}$. Using this, determine a formula for the IMA of both of these wedges in terms of the angle of the wedge $\theta$. Wedge (a) is a right triangle, and wedge (b) is isosceles. Show and/or explain how you got your formula.

Machines QM Wedges.jpg
I agree, this becomes pretty useful in competition. Here's my solution:
$\\IMA=\dfrac{E}{R}\\tan(\frac{\theta}{2})=\frac{R/2}{E}\\2tan(\frac{\theta}{2})=\frac{R}{E}\\\frac{1}{2tan(\frac{\theta}{2})}=\frac{E}{R}\\IMA=\frac{1}{2tan(\frac{\theta}{2})}$
Looks great for wedge B! Wedge A is a little different (not sure if you just missed it or...?)

And yeah, I did the derivation a couple of times and the formula is completely solidified in my head–again would recommend trying this out to anyone who does Machines.
Oh yeah whoops forgot about wedge A
$\\IMA=\dfrac{E}{R}\\tan(\theta)=\frac{R}{E}\\\frac{1}{tan(\theta)}=\frac{E}{R}\\IMA=\frac{1}{tan(\theta)}$

RiverWalker88
Exalted Member
Posts: 103
Joined: February 24th, 2020, 7:14 pm
Division: C
State: NM
Pronouns: He/Him/His
Has thanked: 79 times
Been thanked: 143 times
Contact:

### Re: Machines B/C

Ttonyxx wrote:
March 3rd, 2021, 1:24 pm
RiverWalker88 wrote:
February 25th, 2021, 9:34 pm
Ttonyxx wrote:
February 25th, 2021, 9:12 pm

I agree, this becomes pretty useful in competition. Here's my solution:
$\\IMA=\dfrac{E}{R}\\tan(\frac{\theta}{2})=\frac{R/2}{E}\\2tan(\frac{\theta}{2})=\frac{R}{E}\\\frac{1}{2tan(\frac{\theta}{2})}=\frac{E}{R}\\IMA=\frac{1}{2tan(\frac{\theta}{2})}$
Looks great for wedge B! Wedge A is a little different (not sure if you just missed it or...?)

And yeah, I did the derivation a couple of times and the formula is completely solidified in my head–again would recommend trying this out to anyone who does Machines.
Oh yeah whoops forgot about wedge A
$\\IMA=\dfrac{E}{R}\\tan(\theta)=\frac{R}{E}\\\frac{1}{tan(\theta)}=\frac{E}{R}\\IMA=\frac{1}{tan(\theta)}$