Ford 1102 Paragraph entracion Theory: Torque is a vector quantity that measure the retation effect of a force on an obje

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Ford 1102 Paragraph entracion Theory: Torque is a vector quantity that measure the retation effect of a force on an object that has a fed articulation (pivot. The magnitude is given by the equation below, with Fforce, position vertor between the origin of the force and plot and the angle between rand F. The direction is perpendicular to the plane formed by and F.The rotation direction of the object is indicated an clock W or countershockwis Cow EXTX Fold When the then one force act on an ect, the net turques calculated considering the sign convention, if the turning tendency in CCW, the torque is positive and if the turning tendency is She torque is negative An incont "OW" Procedure: The bacity uses sided by the Uty of Co B Pace Styles see the "Balam Lt Then set "Mans Labels", "Torr MET eports of the bea Voice Semitive
2. Umson Z Suve Spes in the diagram below a few forces are omitted: normal forces from the beam on the objects and the pivot and the gravitational force of the because they will not contribute to the rotational motion of the beam. Also 8=90° and sin 90=1 27=₁₁-1.₂ = 1m x (5kg x2-1m x (10kg x2)=-49Nm Remove the supports and observe the motion of the beam The bear rotates "clockwise" under the effect of a net torque different than zero. -Put the supports back and place a 5 kg brick at 1 m on the left side of the beam and place the 10 kg brigk at 0.5m on the right side of beam. Remove the supports and observe the beam. is the beam moving? 9.8
PARTA Place object on the left side of the beam at distance "t", as indicated in the table below. Calculate, following the example provided below, the distance "t" at which you need to place object mu on the right side of the bearn such as the beam is in rotational equilibrium net torque "COW" sot torque "OW" or net torque "left side" anet torque "right side Mass ma (kg) 20 Notice that in the table, g-9.8m/s and the units (kg, m, m/s) are simplified from the rotational equilibrium equation Check your answer by placing the objects on the beam and removing the supports 20 8 30 10 0.5mx30kg x 9.8m/s² = x 5kg x 9.8m/s Position z (m) 1₂ 0.5 10 0.25 Fus=8'w's 0.5 10 z=20m rm₂g=r.m.g (0.5X20)=(12)(5) ma (kg) 10 20 9 20 Position □ (m) 20
Document Formatting Include below one picture with the balanced beam (no supports) that corresponds to a situation presented in the table. Place the objects me and mu and me on the left and right side of beam and calculate either the unknown location (ra) of an object or the unknown mass (mind the object that will balance the beam. The same rotational equilibriun condition net torque left side" =oet tocque "right side will be used and first row in the table is a solved example for the situation included below: 5mg+5mg=ryme 0.5m x 30kg x 9.8m/s +1.5mx20kg x 9.8m/s = r₂ x 60kg x 9.8m/s² 0.75m Notice that in the table, g-9.8m/s and the units (kg, m, m/s) are simpl rotational equilibrium equation from the
The equilibrium rotation equation has to be used carefully as the objects 1,2, and 3 have different (left or right side) locations. Write both the value and the correct side (left or right) of Mass m. (kg) 20 20 80 50 Position 1 (m) 1.5m left 1.0m left 1.5m left 1m left Mass m. (kg) 30 60 8 30 80 Position 2 (m) 0.50m left 20m right 0.5m right 0.25m right torque "left side" torque right side (0.5)(30)+(1.5)(20)= r₂ x 60 Mass ma (kg) 60 80 20 Position include below one picture with the balanced beam (no supports) that corresponds to a situation presented in the table 0.75 m right 1.75m