A uniform rod AB is 180Cm long and 2kg. It balances horizontally on a single Support when carrying loads of lkg and 3kg at distances 60cm and 120cm respectively from A. Find the position of the support and the magnitude of the downward thrust

Respuesta :

The position of the support is 100cm from A and the magnitude of the downward thrust is 60N.

  • mass of rod = 2Kg, length of rod = 180cm
  • The distance of 1Kg block from A is 60cm
  • The distance of 3Kg block from A is 120cm
  • Suppose the distance of support from A is x cm
  • mass of part of rod on left side of support = 2 * (x / 180)
  • distance of centre of mass of part of rod on left side of support = x / 2 cm
  • mass of part of rod on right side of support = 2 * (180 - x / 180)
  • distance of centre of mass of part of rod on right side of support = (180 - x) / 2 cm
  • The rod balances horizontally, this means net torque is zero.
  • The force that may cause an item to revolve along an axis is measured as torque. In linear kinematics, force is what drives an object's acceleration. Similar to this, an angular acceleration is brought on by torque. As a result, torque may be thought of as the rotating counterpart to force.
  • We know torque = perpendicular distance * force
  • Since net torque around support is zero :
  • torque due to mass on left of support = torque due to mass on right of support
  • torque of 1Kg block + torque of left part of rod = torque of 3Kg block + torque of right part of rod
  • (x - 60)(1*10) + (x/2)((2x/180)*10) = (120 - x)(3*10) + ((180-x)/2)((2*(180-x)/180)*10)
  • On solving for "x" we get, x = 100cm

  • A thrust is a push or a force (mostly reactive). When a system accelerates or pushes mass in one direction, there is a thrust (force) in the other direction that is equally as strong. Thrust is the force that is given to a surface in a direction that is perpendicular to or parallel to the surface. Therefore, thrust is measured in newtons (symbol: N)
  • thrust = sum of forces on support
  • thrust = weight of blocks + weight of rod
  • thrust = (1 * 10) + (3 * 10) + (2 * 10)
  • thrust = 60N

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