text: 1
  You visited us 1 times! Enjoying our articles? Unlock Full Access!
  Question
  
  # In the figure shown below, a horizontal force F is applied on 4 kg block towards left. If the coefficient of friction between the surfaces are 0.5 and 0.4 as shown in the figure. The value of Tension in the rope and force required just to slide the 4 kg block under 2 kg block is (rope is massless and inextensible)
  
  A
  T=90.9 N and F=29 N
  No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
  B
  T=9.09 N and F=14.5 N
  No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
  C
  T=9.09 N and F=29 N
  Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
  D
  T=90.9 N and F=14.5 N
  No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
  Open in App
  Solution
  
  ## The correct option is C T=9.09 N and F=29 NThe FBDs of the blocks are as shown below R1 and R2 are the normal forces acting on the respective blocks. The condition is just to slide, it means in the question it is talking about limiting condition. So, from FBD of 2 kg we have, R1+Tsin37∘=2g=20 R1=20−3T5 ... (1) f1=Tcos37∘ ⇒ f1=4T5 ... (2) ∵ f1=μR1 ⇒ f1=0.5(20−3T5)=10−3T10... (3) From (3) and (2) we get 4T5=10−3T10 4T5+3T10=10 11T10=10⇒T=10011 N ... (4) From (4) in (1) we get R1=20−3×1005×11 =20−5.454 R1=14.546 N Now, from the FBD of 4 kg we get, At equilibrium, R2=R1+4g=14.546+40=54.546 N F=f1+f2 ∵f2=μ2R2=0.4×54.546=21.8 N and from equation (3) f1=10−3×10010×11=7.27 ⇒F=21.8+7.3=29.1 N
  
  Suggest Corrections
  0
  Join BYJU'S Learning Program
  Related Videos
  Warming Up: Playing with Friction
  PHYSICS
  Watch in App
  Explore more
  Join BYJU'S Learning Program