Engineering Mechanics The maximum frictional force, which comes into play, when a body just begins to slide over the surface of the other body, is known as Static friction Dynamic friction Limiting friction Coefficient of friction Static friction Dynamic friction Limiting friction Coefficient of friction ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If a body is acted upon by a number of coplanar non-concurrent forces, it may Move in any one direction rotating about itself All of these Rotate about itself without moving Be completely at rest Move in any one direction rotating about itself All of these Rotate about itself without moving Be completely at rest ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics On the ladder resting on the ground and leaning against a smooth vertical wall, the force of friction will be Perpendicular to the wall at its upper end Zero at its upper end Upwards at its upper end Downwards at its upper end Perpendicular to the wall at its upper end Zero at its upper end Upwards at its upper end Downwards at its upper end ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics For any system of coplanar forces, the condition of equilibrium is that the All of these Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of moments of all the forces about any point should be zero Algebraic sum of the horizontal components of all the forces should be zero All of these Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of moments of all the forces about any point should be zero Algebraic sum of the horizontal components of all the forces should be zero ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resultant of two equal forces ‘P’ making an angle ‘θ’, is given by 2P tanθ/2 2P cotθ/2 2P sinθ/2 2P cosθ/2 2P tanθ/2 2P cotθ/2 2P sinθ/2 2P cosθ/2 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The Cartesian equation of trajectory is (where u = Velocity of projection, α = Angle of projection, and x, y = Co-ordinates of any point on the trajectory after t seconds.) y = (gx²/2u² cos²α) + x. tanα y = (gx²/2u² cos²α) - x. tanα y = x. tanα - (gx²/2u² cos²α) y = x. tanα + (gx²/2u² cos²α) y = (gx²/2u² cos²α) + x. tanα y = (gx²/2u² cos²α) - x. tanα y = x. tanα - (gx²/2u² cos²α) y = x. tanα + (gx²/2u² cos²α) ANSWER DOWNLOAD EXAMIANS APP
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