Weight, Normal Force, and Tension (Multiple Block Interactions)

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Interactions with Weight, Normal Force, and Tension (Multiple Blocks)

Follow equilibrium situations that have multiple box Interaction. Create equilibrium equations to solve problems on a surface or hanging.

Box A on top of box B on the ground in equilibrium

What are the common force magnitudes in this scenario:

  • The weight mAg is a field force on A but equal in magnitude to the contact force (FA-B) which is the push of A down on box B.

All boxes are not moving in equilibrium (ΣF = 0 N in equilibrium)

 

Two boxes A and B stacked free body diagram

Forces on box A with up being positive:

  • Fg: -mAg
  • +FNA

ΣF = FNA – mAg

Forces on A in Box AB Situation
  • ΣF = 0 N in equilibrium

0 = FNA – mAg

FNA  = mAg

This derivation shows that the normal force which is equal to the force B pushes on A (FB-A) is equal to the weight of Box A.

Forces on box B with up being positive:

  • Fg: -mBg
  • FA-B: -mAg
  • +FNB

ΣF = FN – FA-B – mBg

Forces on B in Box AB Situation
  • ΣF = 0 N in equilibrium

0 = FNB – FA-B – mBg

FA-B = FNB – mBg

This derivation shows that the force of A pushing on B is equal to the the normal force pushing up on B minus the weight of box B.

FNB = FA-B + mBg

This derivation shows that the normal force pushing up on B is equal to the force A pushes down on B plus the weight of box B.

Further substitution since magnitude of FA-B = mAg

ΣF = FN – mAg – mBg

  • ΣF = 0 N in equilibrium

0 = FNB – mAg – mBg

FNB = mAg + mBg

This derivation shows that the normal force pushing up on B is equal to the weight of both masses combined.

Two Boxes Hanging and Tension

What are the common force magnitudes in this scenario:

  • B is hanging with a tension up of (FT B-A) which is equal to box B’s pull down on A.
  • Box A is pulling B up by (FT A-B) which is equal and opposite to (FT B-A).

All boxes are in equilibrium (ΣF = 0N in equilibrium)

Two Hanging Boxes Tension

Forces on box A with up being positive:

  • Fg: -mAg
  • -FT B-A
  • +FT

ΣF = FT – mAg – FT B-A

Forces on Hanging Box A
  • ΣF = 0N in equilibrium

0 = FT – mAg - FT B-A

FT = mAg + FT B-A

The tension in the rope above box A is equal in magnitude to the weight of box A plus the tension of box B pulling A down

FT B-A = FT – mAg

The tension in the rope below A is equal to the tension in the rope above A minus the weight of box A.

Forces on box B with up being positive:

  • Fg: -mBg
  • FT A-B

ΣF = FT A-B – mBg

Forces on hanging box B
  • ΣF = 0N in equilibrium

0 = FT A-B – mBg

FT A-B = mBg

The tension in the rope above box B is equal to the magnitude of box B's weight.

Further substitution

  • FT B-A = FT – mAg
  • FT A-B = mBg

Since FT B-A = FT A-B = mBg

mBg = FT – mAg

FT = mAg + mBg

The tension in the rope above box A is equal to the weight of box A and box B combined.

Two Boxes Hanging with Tension and Normal Force

Box A hanging with box B hanging below on the ground in equilibrium

What are the common force magnitudes in this scenario:

  • B is hanging with a tension up of (FT B-A) which is equal to box B’s pull down on A.
  • Box A is pulling B up by (FT A-B) which is equal and opposite to (FT B-A).
    • This is less than the previous scenario because of the ground applying some force up on block B lessening the amount of tension in the rope between block A and B.

***note: my FT A-B subscript represents the force of tension from the block A side of the rope on block B.

Two Hanging Boxes Tension and Normal Force

Forces on box A with up being positive:

  • Fg: -mAg
  • -FT B-A
  • +FT

ΣF = FT – mAg – FT B-A

Box A with Box B on the ground

This scenario is the same as the prior but there will be less pull down as a result of box B on the ground with normal force.

  • ΣF = 0N in equilibrium

0 = FT – mAg - FT B-A

FT = mAg + FT B-A

The tension in the rope above box A is equal in magnitude to the weight of box A plus the tension of box B pulling A down

FT B-A = FT – mAg

The tension in the rope below A is equal to the tension in the rope above A minus the weight of box A.

Forces on box B with up being positive:

  • Fg: -mBg
  • +FT A-B
  • +FNB

ΣF = FT A-B + FNB – mBg

Box B with Box B on the ground
  • ΣF = 0N in equilibrium

0 = FT A-B + FNB – mBg

FT A-B = mBg - FNB

The tension in the rope above box B is equal to the magnitude of box B's weight minus the normal force of the ground pushing up.

FNB = mBg - FT A-B

The normal force of the ground pushing up is equal to the magnitude of the weight of box B minus the tension in the rope above B.

Further substitution:

  • FT B-A = FT – mAg
  • FT A-B = mBg - FNB

FT = mAg + FT B-A

The tension above box A is equal to the magnitude of box A's weight plus the tension in the rope below A.

Since FT B-A = FT A-B = mBg - FNB

FT = mAg + mBg - FNB

FT = g(mA + mB) - FNB

 

After a little more rearranging you see the tension above box A is equal to the combined weight of box A and B minus the normal force below box B.

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