Team Sliders - University of Idaho



Team Sliders

TK Solver Math Analysis

|In-Between Brake System |

|τf=μ1*Pf |

|τr=μ1*Pr |

|tand(θ)=(x1+12)/23 |

|  |

|Upper block |

|Summation forces in the x-direction |

|(F*cosd(270+θ))+(Pf*cosd(θ1))+(Pr*cosd(θ2))+(τf*cosd(θ3))+(τr*cosd(θ4))=0 |

|Summation forces in the y-direction |

|(F*sind(270+θ))+(Pf*sind(θ1))+(Pr*sind(θ2))+(τf*sind(θ3))+(τr*sind(θ4))=0 |

|Summation of moments about point where P and forces cross on rear of block |

|(F*cosd(270+ θ)*7.5)-(F*sind(270+ θ)*(x1+(9+3/8)))-(Pf*sind(θ 1)*18.75)-(τf*sind(θ3)*18.75)=C1 |

|  |

|Lower block |

|Summation forces in the x-direction |

|(F*cosd(90+ θ))+(Pb*cosd(θ5))+(τb*cosd(θ6))+Ff=0 |

|Summation forces in the y-direction |

|(F*sind(90+ θ))+(Pb*sind(θ5))+(τb*sind(θ6))+N=0 |

|Summation of moments about the point where the P and forces cross |

|(-F*cosd(90+ θ)*d)-(F*sind(90+ θ)*x2)+Ff*(l-d)+N*xN=C2 |

|  |

|Ff= μ2*N |

|τb=μ3*Pb |

|θ3= θ1+90 |

|θ4= θ2+90 |

|180- θ1= θ 2 |

|θ6= θ5+90 |

|Input |Name |Output |Unit |Comment |

|  |τf |1431.94590238184 |  |Shear force at front tire |

|.7 |μ1 |  |  |Friction force between the tires and the upper block |

|  |Pf |2045.63700340263 |  |Pressure force on the front tire |

|  |τr |1509.22243720431 |  |Shear force on the rear tire |

|  |Pr |2156.03205314902 |  |Pressure force on the rear tire |

|  |θ |37.59764260401 |  |Angle of the force caused by the bottom block from vertical |

|  |F |2567.049388505 |  |Force from the bottom block |

|30 |θ1 |  |  |Angle for Pf |

|  |θ2 |150 |  |Angle for f |

|  |θ3 |120 |  |Angle for Pr |

|  |θ4 |240 |  |Angle for r |

|  |x1 |5.71087696821313 |  |Distance from the center point of the upper block to the connection point |

|2500 |Pb |  |  |Pressure force on the bottom block from the tire |

|250 |θ5 |  |  |Angle from Pb |

|  |τb |1750 |  |Shear force on the bottom of the tire to the block |

|  |θ6 |340 |  |Angle from b |

|  |Ff |776.777359264322 |  |Friction force from the bottom block to the road |

|  |N |913.855716781555 |  |Normal force on the bottom of the bottom block |

|2 |d |  |  |Vertical distance from the force connection point to the center of the block |

|6 |x2 |  |  |Horizontal distance from the connection point to the center |

|3 |l |  |  |Vertical height from the road to the connection point |

|.85 |μ2 |  |  |friction coefficient between the road and the bottom block |

|0 |C1 |  |  |Constant |

|  |xN |9.07617124868975 |  |horizontal distance from center to the normal force |

|0 |C2 |  |  |Constant |

|.7 |μ3 |  |  |Friction coefficient between the tire and the bottom block |

IN-BETWEEN MODEL CONCLUSION

This model is stating that the forces that are applied to both of the axles will be enough to lock up the tires at any speed, or whether the truck is loaded or empty, when this braking system is applied. Locking up both of the rear axles will be more than enough force required to stop a truck in 800 feet, according to our Stopping Distance Model.

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