Need help on these problems! will thumbs up! U6Q1 A 200 gram mass is placed upon

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Need help on these problems! will thumbs up!

U6Q1 A 200 gram mass is placed upon a balance table at (0,3) it would produce a torque about the A) x-axis B)y-axis C) z- axis U6Q2 Where would a 100 gram mass be placed to bring the balance table into equilibrium? A) (0,3 B (-3,0 C) (0,-6D) (6,0) U6P1 Located on a balance table are the following masses and their positions 100 gm @(0,4), 200gm@(2,-5), 300gm@(3,7) and 400gm(5,0) Determine the total torque about the x and y axis Determine the location of the center of gravity for these masses Where would a 1000 gm mass be placed to bring the system into equilibrium? U6P2 An irregular shaped object has a mass of 325 grams and is placed upon a balance table. The table is brought into equilibrium by placing 2 100 gram masses at (2,0) and (0,3). Where on the balance table is the location of the center of mass for the object?

Explanation / Answer

1) What is torque ????

Torque = force*perpendicular distance

If mass is located on a table ( at 0,3) , it means it lies on the y-axis (if we were to draw x-y axis system on table, we can clearly see this), The weight of the mass will be acting downward. Note that I have used SI units

So this mass will produce a torque of about 5.88 Nm ( Torque = 200/1000*9.8*3) about x-axis . Option a is correct .

2) To balance this 5.88 Nm, we need to place 100 g at (0,-6) . So we have a torque = 100/1000*9.8*(-6) = -5.88 Nm

option c is correct

3) About x-axis = (100*4) + (200 * -5) + (300*7) = 1500

about y axis = (200 *2) + (300*3) + (400*5) = 3300

Center of gravity ( -3.3 , - 1.5)

4)

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