Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

baddinlol Group Title

A block of mass 10kg is pulled along a horizontal plane by a force of 10N inclined at 30* to the plane. The coefficient of friction between the block and the plane is 0.05, Find the acceleration of the block.

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1327647830040:dw|

    • 2 years ago
  2. IsTim Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Wrong group. There is a physics group though. Post your question there.

    • 2 years ago
  3. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    It is in my mathematics books though.... sorry

    • 2 years ago
  4. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    well, F = m*a Force in the horizontal direction is 10*cos(30) = 5sqrt3 not sure where the coefficient of friction goes though...been awhile since physics

    • 2 years ago
  5. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    This question is acceptable here -- don't worry. But you have more like in the physics group.

    • 2 years ago
  6. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Luck, rather.

    • 2 years ago
  7. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Do I subtract the friction coeffecient from the opposing force? I am confused...

    • 2 years ago
  8. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I will try the physics one if it doesn't work out =)

    • 2 years ago
  9. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I believe you need to multiply the normal force by the friction coefficient.

    • 2 years ago
  10. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    And doing that will tell you the frictional force. You add that frictional force as a force vector, if I remember correctly.

    • 2 years ago
  11. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    hmm i think i got it now the vertical force = 10*sin(30) = 5 the downward force = 10g = 98 therefore R = 98 - 5 = 93 R* friction = 93*.05 = 4.65 but thats going in opposite direction net forward force = 5sqrt3 - 4.65 = 4.01 now apply F=ma 4.01 = 10a a = 0.401

    • 2 years ago
  12. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks, that looks right.

    • 2 years ago
  13. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Can I do this in i-j notation?

    • 2 years ago
  14. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i guess, you could write the vectors in i-j notation

    • 2 years ago
  15. baddinlol Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks a lot, really helped

    • 2 years ago
  16. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[F_n=mg \cos \theta =(10kg)(9.81\frac{N}{kg})(\cos(30 \deg))=85.N\]\[F_f = μN\]\[F_f=(0.05)(85.N)=4.2N\]\[F=85.N-4.2N=80.8N\]\[F=ma\]\[80.8N=(10kg)a \rightarrow a=8.08\frac{N}{kg}=8.08\frac{m}{s^2}\]This is what I get. I don't think you found the force normal correctly, crow.

    • 2 years ago
  17. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    oh oops, i thought that the Net force had to be 0 downward force, mg = 98N so upward force must be 98N, no?

    • 2 years ago
  18. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If the net force were 0, the block wouldn't be accelerating would it?

    • 2 years ago
  19. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    no i was speaking of vertical forces, the block is staying on the ground. isn't the only Force pushing it forward is the 10N applied at the 30* angle ?

    • 2 years ago
  20. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, and so it wouldn't make sense for the normal force to be more than 10. Is the 10N even enough to overcome gravity? \[F_n=mgcos \theta=(10kg)(-9.81\frac{N}{kg})(\cos(30\deg))=-85.0N\]

    • 2 years ago
  21. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i dunno, that doesn't seem right the normal force is pointing straight up because the block is moving on a horizontal plane so you don't need to multiply mg by cos* the normal force will just be enough to cancel gravity -> Fn = mg - 10sin(3)

    • 2 years ago
  22. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh! I misread this question, and the picture only confused me more. I was thinking of it as a 30degree incline. Geez. Ok.

    • 2 years ago
  23. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    ahh, yeah so you're right if the block is moving up the incline

    • 2 years ago
  24. Xishem Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep, and your old solution is correct :P

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.