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Math Help - Mechanics rigid objects in equilibrium

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    Mechanics rigid objects in equilibrium

    Mechanics rigid objects in equilibrium-pic_0901.jpgMechanics rigid objects in equilibrium-pic_0871.jpg answer for first part is 0.275
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    MHF Contributor ebaines's Avatar
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    Re: Mechanics rigid objects in equilibrium

    Yes, the answer is 0.275. Do you have a question?
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    Re: Mechanics rigid objects in equilibrium

    i dont get how the answer is 0.275 can u help me get to that i just showed my approach to this question but yet i cant get it
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    Re: Mechanics rigid objects in equilibrium

    Quote Originally Posted by ebaines View Post
    Yes, the answer is 0.275. Do you have a question?
    help i dont get it
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    MHF Contributor ebaines's Avatar
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    Re: Mechanics rigid objects in equilibrium

    Step 1: Determine the normal force of the rear leg to the ground. You can take sum of torques about the front tire contact point and set it to zero. Remember that the torque due to the weight is  \vec T = \vec R \bold {x} \vec W , where  \vec R is the vector from the tire contact point to the center of mass.

    Step 2: Determine the friction force of the rear leg, from sum of forces parallel to the incline = 0.

    Step 3: coefficient of static friction = friction force divided by the normal force of the leg.

    If you get stuck post back with your attempt and we can check it for you.
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    Re: Mechanics rigid objects in equilibrium

    Quote Originally Posted by ebaines View Post
    Step 1: Determine the normal force of the rear leg to the ground. You can take sum of torques about the front tire contact point and set it to zero. Remember that the torque due to the weight is  \vec T = \vec R \bold {x} \vec W , where  \vec R is the vector from the tire contact point to the center of mass.

    Step 2: Determine the friction force of the rear leg, from sum of forces parallel to the incline = 0.

    Step 3: coefficient of static friction = friction force divided by the normal force of the leg.

    If you get stuck post back with your attempt and we can check it for you.
    i just did that u can check one of the two pics that i uploaded and i dont know how to apply cross products in this moment question
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  7. #7
    MHF Contributor ebaines's Avatar
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    Re: Mechanics rigid objects in equilibrium

    This part is wrong:

     R \times 120 = (300 \cos(6) \times 40)+(300 \sin(6) \times 50)

    Two issues:

    1. When you break the weight into these two components one acts in a clockwise direction about the tire and the other acts in a counter-clockwise fashion - hence the term involving the 40cm moment arm should be negative.
    2. You have switched cosine and sine terms. So it should be:

     R \times 120 = (300 \cos(6) \times 50) - (300 \sin(6) \times 40)

    Try it with these corrections.
    Last edited by ebaines; March 10th 2014 at 10:39 AM.
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    Re: Mechanics rigid objects in equilibrium [Solved]

    for second part i did (100*300cos(6))-(60-40)*300sin(6)=150*R
    N+R=300cos(6) where N is the perpendicular to the path solving this i get N=103.6 answer is 104 did i get things right?
    Last edited by abdulrehmanshah; March 10th 2014 at 08:23 PM.
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  9. #9
    MHF Contributor ebaines's Avatar
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    Re: Mechanics rigid objects in equilibrium [Solved]

    Quote Originally Posted by abdulrehmanshah View Post
    i get N=103.6 answer is 104 did i get things right?
    Yes, that's the force the gardeener applies perpendicular to the path. Next you need to determine the horizontal component.
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