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Homework Help: General questions

  1. Nov 20, 2007 #1
    1). A uniform bridge span weights 50x10^3 N and is 40m long. An automobile weighting 15x10^3 N is parked with its center of gravity located 12m from the right pier. WHat upward force does the left pier provide? The answer is 29.5x10^3 N but how do I get that?
    my attempt: F - (50000N)(20m) - (15000N)(8m)=0

    2). Tasha has a mass of 20kg and wants to use a 4m board of mass 10kg as a seesaw. Her friends are too busy, so Tasha seesaws alone by putting the support at the systems center of gravity, she sits on one end of the board. How far from the support point is she? The answer is .67m, but how is this problem done?
    my atempt: -(10kg)(2m)- (20kg)(x)(9.8)=0

    3). A 100N uniiform ladder, 8m long rests against a smooth vertical wall. The coefficient of static friction between the ladder andthe floor is 0.40. What minimum angle can the ladder make with the floor before it slips? THe answer is 51'
    my attempt: NO CLUE

    4). A disk has a moment of inertia of 3x10^-4 kg.m2 and rotates with an angular speed of 3.5rads/sec. What net torque must be applied to bring it to rest within 3 seconds?
    The answer is 3.5x10^-4 N.m but how do they get this answer?
    my attempt: I know that the I is very small so we can ignore it

    5). A buket of water with a total mass of 23kg is attatched to a rope, which in turn is wound around a 0.050m radius cylinder at thetop of a well. A crank with a turning raidus of 0.25m is attached to the end of the cylinder and the moment of inertia of the cylinder and the crank is 0.12kg.m2. If the bucket is raised to the top of the well and released, what is the acceleration of the bucket as it falls toward the bottom of the well?
    The answer is 3.2m/s2 but how?
    my attempt: NO CLUE
  2. jcsd
  3. Nov 20, 2007 #2


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    You cannot do any problem in mechanics without having a picture of what is happening and where the forces are.

    You also need to know that the moment = force * distance (at right angles to force)
    if a see-saw is not moving then the moment on both sides balances.

    1, Draw the diagram with the car in the middle, show the forces and work out the force n each pier. Now draw the forces with the car ontop of the right pier.
    Think about what happens as the car moves toward the 12m point (hint think ratios)

    2, For the girl to see-saw herself then her+part of plank on her side must balnace part of plank on other side.

    3, The sideways force needed to overcome friction is coeff_friction * normal force.
    Normal force is the force downward on the sliding part.

    4, You can't ignore I - it's a multiplication. If I were negligible it would take no torque to bring it to a stop instantly.

    5, F = m a. The acceleration downward on the bucket is g (9.8m/s^2)
    What force is needed to turn the crank ?
  4. Nov 20, 2007 #3
    1. I have drawn this diagram and still no clue. I have played with the numbers for DAYS.
    2. I understand that part
    4. I GOT it:) Tnet=(I)(speed) / time
    5. ?
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