2 non-rectangular blocks, friction, only gravity

In summary, the conversation is about a homework problem involving two blocks and their forces. The equations and attempts at solving the problem are discussed, including the simplification of friction forces and the use of trigonometry. The person is asking for assistance in finding where they may have made a mistake in their solution.
  • #1
trogtothedor
26
0

Homework Statement



An attachment with the problem specifications, HW6 - C, is attached. Here is a link in case that is easier to view.

http://www.flickr.com/photos/54849943@N04/5082480364/

agravity = 32.2 ft/s2

Homework Equations


[tex]\Sigma[/tex]FyA = mA*aA
[tex]\Sigma[/tex]FxA = 0
[tex]\Sigma[/tex]FyB= 0
[tex]\Sigma[/tex]FxB= mB * aB


The Attempt at a Solution


FA on B = FB on A

Writing my force equations, I get:
[tex]\Sigma[/tex]FyA = mA*aA = Ffriction,wall+ Ffriction,B on Asin(70) + FB on A*sin(20)

[tex]\Sigma[/tex]FxA = 0 = Fwall + F friction, B on A*cos(70) - F B on A*cos(20)

[tex]\Sigma[/tex]FyB = 0 = N - 100 -FA on B *sin(20) - Ffriction, A on B *sin(70)

[tex]\Sigma[/tex]FyB = mB * aB = FA on B *cos(20) - F friction, ground - F friction, A on B* cos(70)

Also, to simplify,

Ffriction,wall = Fwall * [tex]\mu[/tex]k

Ffriction, ground = N* [tex]\mu[/tex]k

For both cases, Ffriction, A on B = FA on B* [tex]\mu[/tex]k

Also, since both blocks start from rest, 1 foot = (1/2) * aA * (t2)

Using trigonometry, I also concluded that when block A moves one foot down, block B will have moved 1/tan(70) feet to the right. Thus:

t2 = 2/(aA) = 2/(aB * tan(70))

Am I wrong in any of this? Either something here is wrong, or my algebra is going haywire throughout the solving process.

Thank you in advance for any assistance!
 

Attachments

  • HW6 - C.jpg
    HW6 - C.jpg
    19.5 KB · Views: 400
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  • #2
I anyone can see any area where I've made a mistake, I'd appreciate it! I feel like I covered everything and have all the angles right, but I'm still coming up with the wrong answer.
 
  • #3
I got it. In my coorrdinate system, acceleration A is negative, so -aA=aBtan(70)
 

Related to 2 non-rectangular blocks, friction, only gravity

1. What is the difference between a rectangular and non-rectangular block in terms of friction?

The main difference between a rectangular and non-rectangular block is the surface area in contact with the ground. A rectangular block has more surface area in contact with the ground, which results in higher friction compared to a non-rectangular block.

2. How does the shape of a block affect the amount of friction?

The shape of a block affects the amount of friction by changing the surface area in contact with the ground. A block with a larger surface area in contact with the ground will have higher friction compared to a block with a smaller surface area.

3. Can a non-rectangular block have no friction?

No, a non-rectangular block will always have some friction because there will always be some surface area in contact with the ground. However, the amount of friction may be lower compared to a rectangular block with the same mass and material.

4. How does friction affect the movement of non-rectangular blocks?

Friction acts in the opposite direction of movement, so it will slow down the movement of non-rectangular blocks. The amount of friction will depend on the shape and surface area of the block, as well as the surface it is sliding on.

5. Is gravity the only force affecting the movement of non-rectangular blocks?

No, other forces such as air resistance or applied forces can also affect the movement of non-rectangular blocks. However, in the scenario of only gravity acting on the blocks, the shape will still play a role in the amount of friction and subsequently, the movement of the blocks.

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