In the Fig above, MN is a light uniform meter rule pivoted at O, the 80cm mark. A load of mass 3.00kg is suspended on the meter rule at L, the 10cm mark. If the rule is kept in equilibrium by a string RP, fixed at P and attached to the rule at R, the 20cm mark, then the Tension T on the string is?
3.0N
3.4N
3.5N
6.0N
7.0N
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (17)
Answer to this question should be 70N.
30(70) = T sin 30 (60)
2100 = 30T
T = 70N
The answer should be E because what we have to do is to take moment at the knife edge we should also note that taking the moment at the same line of force for the horizontal component of T = 0
Solution:
Tsin30 × 60cm = 3×70cm
Therefore T = 7kg
It's supposed to be 70N. Tension is a force, therefore the unit is Newton (N). The mass 3kg is supposed to be converted to weight, which would be 30N. The cm converts to m too. Both sides have to be dimensionally equal.
30N × 0.7m = Tsin30° × 0.6m
T = 70N
The answer is not in the options
The question asks for tension. Tension is a type of force. The unit should be Newton not kg. The answer is 70N, not 7N or 7kg.
The answer is 7N. Please endeavour to correct this as there is enough evidence in the comment section.
