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Published byMerilyn Weaver Modified over 9 years ago
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Physics Chp 8
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Torque This is the combination of force applied and distance from an axis point and can cause rotation. The force and distance from the axis must be perpendicular.
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If the torque can cause counter clockwise rotation then it’s positive Clockwise is negative
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The force and distance have to be perpendicular. If at an angle use the angle and sine or cosine to get only the component that is perpendicular! Units are Nm
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What is the net torque on a 4 m teeter totter if you, 58 kg, and your friend, 45 kg, are on opposite ends with the fulcrum (pivot) in the center?
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τ =(58kg)(9.8m/s 2 )(2m) –(45kg)(9.8m/s 2 )(2m) τ = 255 Nm 0 m 2m 58 kg45 kg
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Center of mass Moment of Inertia – Tendency to try to stay rotating – Different for each object as well as how it rotates – Ex. A ball around it’s edge vs it’s center
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In equilibrium ΣF x = 0 Σ F y = 0 Σ τ = 0
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Steps to solve 1) Free-body diagram 2) Choose convenient axis, pivot point 3) Apply the force equations 4) Apply torque equation 5) Solve
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If you, 55 kg, climb a 6.0 m, 25 kg ladder and stand on the rung that is 5.6 m up and the ladder is 40 o from the ground. How much force does the wall apply to the ladder assuming no friction with the wall? How about the ground?
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Fw Fyou Flad Fgy Fgx 6.0m 5.6m 3.0m 40 o
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ΣFx = Fgx – Fw = 0 Fgx = Fw ΣFy = Fgy – Flad – Fyou = 0 Fgy = Flad + Fyou = Fgy = (25kg)(9.8m/s 2 ) + (55kg)(9.8m/s 2 ) Fgy = 784 N
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Στ = Fw(6.0m)sin40 o – Fyou(5.6m)sin50 o – Flad(3.0m)sin50 o = 0 Fw(6.0m)sin40 o = (55kg)(9.8m/s 2 )(5.6m) sin50 o + (25kg)(9.8m/s 2 )(3.0m) sin50 o Fw = 750 N Fgx = Fw = 750 N
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Simple Machines Changes the force applied to an object or the distance it’s moved (or both) Work remains the same (ignoring friction)
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Lever Inclined Plane
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Wheel and Axle
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Wedge
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Pulleys
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Screw Just an incline plane wrapped around itself
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All machines provide a mechanical advantage or why use them (not Rube Goldberg Machines)
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Real Machines Friction causes a loss in the energy to do work Efficiency
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