James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
Hi, me again with another mechanics problem (my favourite):
'A uniform rod AB of weight 80N rests with its lower end A on a rough horizontal floor. A string attached to end B keeps the rod in equilibrium. The string is held at 90 degrees to the rod. The tension in the string is T. The coefficient of friction between the ground and the rod is mu. R is the normal reaction at A and F is the frictional force at A.
Find the magnitudes of T, R and F, and the least possible value of mu.'
The rod is in equilibrium. You are told the rod is uniform, so you can assume the weight acts at the middle. So now even though we are not told the length of AB, we can take moments by writing
Ta = 80 x .5a x cos 30
so T = 40cos30
T = 35N
so far so good.
Now it falls apart.
To work out the other values I resolve the forces vertically and horizontally:
(V) R +T sin30 = 80 .................R + Tcos(30) = 80 ..........................edited
R = 80 - 17.5 = 62.5N (wrong)
(H) F = T cos30 ..........................F = T sin(30) ..........................edited
F = 35 * .866 =30 (also wrong)
Am I missing something?
'A uniform rod AB of weight 80N rests with its lower end A on a rough horizontal floor. A string attached to end B keeps the rod in equilibrium. The string is held at 90 degrees to the rod. The tension in the string is T. The coefficient of friction between the ground and the rod is mu. R is the normal reaction at A and F is the frictional force at A.
Find the magnitudes of T, R and F, and the least possible value of mu.'
The rod is in equilibrium. You are told the rod is uniform, so you can assume the weight acts at the middle. So now even though we are not told the length of AB, we can take moments by writing
Ta = 80 x .5a x cos 30
so T = 40cos30
T = 35N
so far so good.
Now it falls apart.
To work out the other values I resolve the forces vertically and horizontally:
(V) R +
R = 80 - 17.5 = 62.5N (wrong)
(H) F =
F = 35 * .866 =30 (also wrong)
Am I missing something?
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