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Homework Help: Statics practice problems

  1. Nov 18, 2004 #1
    Hi I have a Physics re-test tomorrow on Statics. Since everyone failed the 1st statics test he is going to let us take another one (he's really nice :smile: ). It is not going to be like the 1st one. Where can I get some statics practice problems?? I'm doing ALL of the problems in the book but 95% of the time he does not put problems on the test like the 1s in the book (that's one of the reasons I did not do so well on the 1st test).

  2. jcsd
  3. Nov 18, 2004 #2
    What level physics are you in?
  4. Nov 18, 2004 #3
    AP Physics B
  5. Nov 18, 2004 #4
    Okay straight from my physics B statics test:

    1. A uniform 20.0 kg, 10.0m long beam is supported by two posts. Post A is 1.00m from an end of the beam and post B is 4.00m from the other end. A 15.00 kg cat sits on the beam directly above post B.
    a)Calculate the force exerted by post A on the beam
    b)Calculate force exerted by post B on the beam.
    c)How far can the cat walk toward the end of the board before the board tips. The beam is not fastened to the posts.

    2. A 50.0 kg uniform sign is hung from a 20.0 kg, 4.00m long uniform beam supported by a wall and a wire. The sign is a square 3.00m on a side and is hung so that the end of the sign is even with the end of the beam. The wire makes an angle of 30 degrees with the wall.
    a)Find the tension in the wire.
    b)Find the force exerted by the wall on the beam.

    3. A 5.00m long 20.0 kg uniform ladder leans against a wall so that the top of the ladder is 4.00 m above the ground. Assume no friction between wall and ladder.
    a) What force is exerted by the wall on the ladder?
    b) What force is exerted by the ground on the ladder?
    c)What minimum coefficient of friction between ground and ladder is required to keep the system static.
    Last edited: Nov 18, 2004
  6. Nov 18, 2004 #5
    a.) (20.9g)(4)+(50g)(3)+8(t)=0
    b.) 286.16+ (20.9g)+(50g)+fwall = 0
    fwall= 980.98
    Last edited: Nov 18, 2004
  7. Nov 18, 2004 #6
    Sorry typed some wrong numbers, look at my editted post. Anyway you're not doing the problem right, I'll get back to you on that in a second.
  8. Nov 18, 2004 #7
    a) (Fwall)(4.99)+(5)(20.9g)=0
    Fwall = 205.23
    b) 205.23+(20.9g) + fn = 0
    c) i really don't have a clue
  9. Nov 18, 2004 #8
    Ok Torque counterclockwise = Torque clockwise. Is this what you're trying to do?

    Pick a point to sum your torques around. I'd suggest the point where the beam touches the wall, since you don't yet know the force that the wall exerts on the beam. The beams weight acts down from the center of the beam. You have the center of mass of the sign right. The wire isn't perpendicular to the beam so only the vertical component of the tension is going to be producing torque. Remember Toque is the cross product of force and the radius vectors.
  10. Nov 18, 2004 #9
    Note that the wire is attached to the beam's end opposite the wall.
  11. Nov 18, 2004 #10

    Would I do:

    (20g)(2) +(50g)(3)+ 8T = 0
    t= 232.75
  12. Nov 18, 2004 #11
    Sorry for not reading your first post correctly but 3m isn't the cm of the sign and where is the 8 coming from?
  13. Nov 18, 2004 #12
    Woops that is suppose to be 40. I did the math wrong. That is the length of the wire.

    cos 60 = 20/h
    h = 40

    So would the answer be

    (20g)(2) +(50g)(1.5)+ 40T = 0
    t= 28.175
  14. Nov 18, 2004 #13
    Torque= the cross product of F and R
    Cross product is always F*R * sine of the lesser included angle.
    So the torque from the wire is [tex] Tsin(60) * 4m [/tex] The length of the wire is irrevelant here.
  15. Nov 18, 2004 #14
    Is T the force?? If so would T be 20*9.8?
    Last edited: Nov 18, 2004
  16. Nov 18, 2004 #15
    what would you do with the 50 kg block?
  17. Nov 18, 2004 #16
    2(20g)+ (4)(50g) = 4Tsin 60
    t= 678.96
  18. Nov 18, 2004 #17
    Very close. You have everything correct except the radius for the sign. It's 3m long so where is its center of mass. How far away from the center of rotation is the center of mass?
  19. Nov 18, 2004 #18
    It's center of mass is at 1.5 right? So what would I do with the 1.5?
  20. Nov 18, 2004 #19
    The center of mass is 1.5m away from the edge of the sign as you stated. The sign is attached one end to the 4m mark and one end to the 1m mark.
  21. Nov 18, 2004 #20
    2(20g)+ (4)(50g) + (5.5)(50g)= 4Tsin 60
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