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Blocks A and B of Fig. have 20- and 50-kg masses, respectively. The static coefficients of friction be- tween blocks A and B and between block B and the hori- zontal surface are 0.25. The static coefficient of friction be- tween the rope and the fixed pegs is 0.20. Determine the force P required to initiate motion of block B. P B

Blocks A and B of Fig. have 20- and 50-kg masses, respectively. The static coefficients of friction be- tween blocks A and B and between block B and the hori- zontal surface are 0.25. The static coefficient of friction be- tween the rope and the fixed pegs is 0.20. Determine the force P required to initiate motion of block B. P B

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Blocks A and B of Fig. have 20- and 50-kg masses, respectively. The static coefficients of friction be- tween blocks A and B and between block B and the hori- zontal surface are 0.25. The static coefficient of friction be- tween the rope and the fixed pegs is 0.20. Determine the force P required to initiate motion of block B. P B
Transcribed Image Text:Blocks A and B of Fig. have 20- and 50-kg masses, respectively. The static coefficients of friction be- tween blocks A and B and between block B and the hori- zontal surface are 0.25. The static coefficient of friction be- tween the rope and the fixed pegs is 0.20. Determine the force P required to initiate motion of block B. P B
Expert Solution
Step 1

Given data:

  • The mass of the block A is mA=20 kg.
  • The mass of the block B is mB=50 kg.
  • The coefficient of static friction between block A and B is μ1=0.25.
  • The coefficient of static friction between rope and fixed pegs is μ2=0.2.

First draw the FBD for block A as:

Advanced Physics homework question answer, step 1, image 1

Now apply the force equilibrium condition along y-direction as:

∑Fy=0NA-mAg=0NA-20 kg×9.81 m/s2×1 Nkg·ms2=0NA=196.2 N

For impending slipping:

fA=μ1NAfA=0.25×196.2 NfA=49.05 N

Step 2

Draw the FBD for block B as:

Advanced Physics homework question answer, step 2, image 1

Now apply the force equilibrium condition along y-direction as:

∑Fy=0NB-NA-mBg=0NB-196.2 N-50 kg×9.81 m/s2×1 Nkg·ms2=0NB=686.7 N

For impending slipping:

fB=μ1NBfB=0.25×686.7 NfB=171.675 N

Now apply the force equilibrium condition along x-direction as:

∑Fx=0fA+fB-T1=049.05 N+171.675 N-T1=0T1=220.725 N

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