A bar of length L, 50 cm, is attached to a wall and is free to rotate. A mass of

Question

A bar of length L, 50 cm, is attached to a wall and is free to rotate. A mass of 3kg is attached to the very end of the bar. A string runs from the wall to the bar with the conditions listed below.

A) What is the tension of the string if the bar is massless?
B) Repeat part A, but the bar now has a linear density of 25 g/m. (x is the distance from the left of the bar, x needs to be in meters.)
C) You have a massless bar. Hanging from the bar are masses which are listed below. Where do you place the fulcrum such that the system is balanced?
D) Repeat part C, but the bar now has a linear density of 35 g/m. (x is the distance from the left of the bar, x needs to be in meters.)

Explanation / Answer

50cm 3kg

Find the combined vectors in the string. The force due to gravity(weight) of the object is F=mg m=mass g=acceleration due to gravity(9.805m/s2). Once you have that, get the angle between the bar and the point at which the string is placed in the wall. This will allow you to find the vectors in the y-dimension(which is the weight of the object=F) and the x-dimension(Fcos()). To solve for B) just multiply the density of the bar times the lenght of the bar(in meters). For the rest of the answers I need more info. but it's the same process.

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