1arey waha pe tension calculate karte hain naa...
v = root(T/mu) ...
16 Answers
21BTW ye mera dbt nahi hai [3]
par ha i thot differently,
@ Asis can u posst ur methd i think its diff frm wat i did by takin normal reaction force and its components
106arey mujhe abhi woh tuition copy dhoondnaa padega... usi mein likhaa tha aur usi type ke aur kuch probs diye the ... jab mielga tab post kar doonga .. abhi mujhe woh yaad nahiin hai....
33W=Ncos(300),
N=2W/√3
Nsin(300)=W/√3
F=Wdl/√32Ï€r
F=Wrdθ/√32Ï€r .....r=R/2
F=2Tsin(dθ/2)
F=2Tdθ/2=Tdθ
Wrdθ/√32Ï€r=Tdθ
T=W/2√3Ï€ ...(ans)
33If coefficient of friction would have been μ
then..
T=K(sinθ-μcosθ)/(cosθ+μsinθ)
bcause for μ=tanθ T should be zero.. [3] by my tanθ theorem... [3]
and in denominator...just opposite... bcause...when we take component of any force N and μN is two perp directions then once Nsinθ then Ncosθ.. [3]
But earlier when θ=300 and μ=0 we have the answer... so K=mg/2π..(in between just T=K(..) can help you to match the options..
[3]
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This was tukka..
Now just posting the solution for friction coefficient=μ
33Ncosθ+μNsinθ=mg
Nsinθ-μNcosθ)=F'
F'rdθ/2πr=2Tdθ/2
N(sinθ-μNcosθ)dθ/2π=Tdθ
Putting N from 1st equaion
N=mg/(cosθ+μsinθ)..
\frac{mg}{2\pi}(\frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta })
21arre kuch nahi.....
USKA apna theorem hai [4]
matlab its jus a deduction
33:D
tanθ theorem...
Whenever q is related with friction answer is almost always tanθ....
(given you choose θ correctly... else it would come out to be cotθ.)
[3]