Technical Physics: Mechanics You may work with physics classmates & use our cour

Question

Technical Physics: Mechanics You may work with physics classmates & use our course materials (handouts, book. labs, to complete this qui due at the start of our next class Name: Jonah stupar 1. Short answer: 1 each for a total of 4 points) i. Why do we need both sum of the torques the sum of the forces to equal zero for a the as well as system to be in equilibrium (static)? Explain using physics ii. In the figure to the right, four (4) boxes are shown with their centers of mass indicated by a solid dot. In which of the four orientations shown, if any, does the box tip over? Explain sin iii. In the diagram below, what is the mass of the rod if this system is in balance? Show support for your answer using physics (such as calculations)! 8 m 5kg Page 1 of 4

Explanation / Answer

i)

When we have a number of forces F1,F2,F3, ..... Fn acting on a body. All the forces vecorially addup to a single effctive force say Feff, Now if the magnitude of this effective force is no-zero then the body is subjected to an acceleration as per newtons second law

F = ma and the body will be in motion and cannot be in equilibrium position, hence the sum of the forces must be 0 for equilibrium.

When the sum of the forces is zero, the point of action of all the forces may not be unique. In that case each force will produce a torque on the object which is equal to r X F.

When we add up all the troques acting on the body it must be 0 if not the body will have a rotational motion .

ii) case c: In ths case the vertical line from the COM, thw eight vector falls outside the left edge of the box and produces a net CCW torque about the point of contact which may trip the box.

iii) 5Kg, the rod weight acts from the center i.e. 4 m from the edge, which is 2m from the point of support. The mass suspended at one edge is 5Kg and is 2m from the support, both the weights shall produce equal and opposite torques about the point of support for equilibrium.

iv) Fk , frictional force, only this force has a non-zero torque arm, all other force lines pass through the center and the arm lenght is o, hence 0 torque.

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