#1

A uniform rod of length 60cm has a weight of 14N. It is pivoted at one end and held in a horizontal position by a thread tied to its other end. The thread makes an angle of 50 degrees with the horizontal. Calculate:
1-the moment of the weight of the rod about the pivot
2-the tension T in the thread required to hold the rod horizontally.

--------
moment of force = force * perpendicular distance from pivot
The answer is 4.2 Nm = 0.3m * 14N
But why is the perpendicular distance 0.3m shouldn't it be 0.6m? Is it because you measure from the centre of gravity of the rod?
-------
For part 2 i don't have a clue. Can someone explain it to me please?
0
2 years ago
#2
[QUOTE=zattyzatzat;71331930]
A uniform rod of length 60cm has a weight of 14N. It is pivoted at one end and held in a horizontal position by a thread tied to its other end. The thread makes an angle of 50 degrees with the horizontal. Calculate:
1-the moment of the weight of the rod about the pivot
2-the tension T in the thread required to hold the rod horizontally.

--------
1. moment of force = force * perpendicular distance from pivot

YES the moment acts through the center of gravity of the rod at 0.30m from pivot end:
Mrod = 14 X 0.30 = 4.2Nm (clockkwise)

2. The tension on the thread must provide a balancing moment in opposite sense for static equilibrium of the rod:
Mthread = (Tsin50o) X 0.60 = - 4.2Nm (anticlockwise)
T = 4.2/[(sin50o) X 0.60] = 9.14N

(we take moments about any point really but usually use pivots so we don't have to work out reaction forces there!)
0
#3
ah thanks for clearing that up
0
2 years ago
#4
(Original post by zattyzatzat)
ah thanks for clearing that up
No problem what course are you doing?
0
#5
Cambridge International Physics A-Level
0
2 years ago
#6
(Original post by zattyzatzat)
Cambridge International Physics A-Level
Might have been a good idea to rename thread title as
"Cambridge Intl Physics A-level 2017 discussion thread"
but that's upto you- if you want to do that just ask the mods.

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