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New answer posted
3 weeks agoContributor-Level 9
Acceleration of the combined system along the incline is given by
With respect to block M :

New answer posted
3 weeks agoContributor-Level 10
I? = P? – P? = 3 * 4? – 3 * 3î = 12? – 9î |I? | = √12² + 9² = 15kgm/s
New answer posted
3 weeks agoContributor-Level 10
N = (mg/√2) + (mω²h/√2)
(mω²h/√2) = (mg/√2) + f?
(mω²h/√2) = (mg/√2) + μ (mg/√2) + (mω²h/√2)
ω = √ (g/h) (1+μ)/ (1-μ) = 10rad/s
New answer posted
3 weeks agoContributor-Level 10
Let the acceleration of string = a

For 2 kg 20 – T1 = 2a ………… (i)
For monkey T2 – 80 – T1 = 8 (2 – a) …. (ii)
For 10 kg T2 – 100 = 10a ……. (iii)
Acceleration of monkey w.r. to ground
= 2 – 0.8 = 1.2 m/s2
t = 2 sec
New answer posted
3 weeks agoContributor-Level 10
Centripetal acceleration is provided by friction force. Maximum friction force should be equal to or greater that mv² / R.
µmg = mv²/R ; µ = 100/ (100*10) ; µ = 1/10 ; µ = 0.1
New answer posted
3 weeks agoContributor-Level 10
Taking torque at top pt, mg sin30°*L / 2 = F cos30°*L ⇒ F = mg / 2√3
New answer posted
3 weeks agoContributor-Level 10
Applying conservation of momentum, m (ui + 2uj) + 3m (0) = 4m (vi +vj)
⇒ v = u / 4 and v' = u / 2
New answer posted
3 weeks agoContributor-Level 9
F.B.D. of hanging length
F.B.D. of chain lying on the table
f = T =
x = 2m

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