Nįor the rod to be in translational equilibrium horizontally, the horizontal component of The vertical component of □ is □ □ c o s, soĪnd it follows that the magnitude of □ is given by Which shows that the magnitude of □ is given byįor the rod to be in vertical translational equilibrium, the magnitude of the weight must equal the sum of c o s s i n Īnd c o s □ = 4 5, this can be rewritten as The rod is in equilibrium, so the counterclockwise moments can be equated to the clockwise moments:Ħ. Then the direction of the reaction force on the object is perpendicular to the inclination of the object. If an extended object that is itself inclined at some angle is supported at a point, This force is called the normal reaction. Is perpendicular to the direction of the surface. When an object is on an inclined surface, the direction of the reaction force on the object The magnitudes and directions of reaction forces on rigid bodies can be determined from the type ofĬontact the body has with the source of the reaction forces. Which determines the magnitude of force □, which is given by Equating the counterclockwise and clockwise moments givesĢ. 1 □ □ c o s and the counterclockwise moments about □ The counterclockwise moments on the rod must therefore be equal to the clockwise moments on it. Since the rod is in rotational equilibrium, the net moment on it must be zero. For the unknown force, this angle is not stated,īut it is the co-angle of angle □ and so it is equal to □. Because these forces pass through point □, they produce zero moment about □.Įach of the applied forces has a horizontal component equal to the cosine of the angle from the Is the weight of the rod, □ is the reaction at □,Īnd □ is the force due to friction. The following figure shows the forces that should be included, where □ Net horizontal force acting on the rod, and it acts in the opposite direction to The force due to friction for an object in equilibrium is equal in magnitude to the The missing force is the forceĭue to friction between the base of the rod and the surface that the rod rests on. Must exist that is not included in the original figure. There would only be one nonzero moment acting on the rod. It would also initially seem that the rod cannot be in rotational equilibrium,īecause if moments were taken about either of the points where an applied force acts, Must be equal to the sum of the weight and the vertically downward components of the applied forces. The net vertical force acting on it must be zero, so the reaction force at □ Since the rod is in translational equilibrium, Is the reaction force from the surface at □. The weight of the rod is not shown and neither The figure does notĪctually show all the forces acting on the rod. The entire equation of motion is F=Kx +Cv+Ma if velocity and acceleration has become zero than F=Kx, and the stress in the body obeys Hooke's Law.It would initially seem that the rod cannot be in translational equilibrium,Īs both forces acting on it have downward vertical components. OR rigid means it is in static equilibrium which means its velocity and acceleration is zero.Īn object in static equilibrium is unable to move because all the forces acting on it compensate for one another, that doesn't mean the body isn't strained and deformed. It isn't clear from your post if rigid means in-compressible, zero strains. Once it is a rest on the moon, it is still deformed and has strains. It under goes velocities, rigid body translation, accelerations, etc. Then I can take that stressed cube and put it on a rocket to the moon. For example I can compress a cube on earth and it is deformed, has strains and stresses. A body can have a rigid body motions, be in static equilibrium, have zero velocity, and acceleration, and still under strains a deformations within the body that produces stresses. You seem to be mixing the terms motion and deformation.
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