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Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by...

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F⃗ . (Figure 1) The magnitude of the tension in the string between blocks B and C is T = 3.00 N . Assume that each block has mass m = 0.400 kg .

a) What is the magnitude F of the force?

b)What is the tension TAB in the string between block A and block B?

Solutions

Expert Solution

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F. The magnitude of the tension in the string between blocks B and C is T= 3.00N . Assume that each block has mass m= 0.400kg.

The force, F, is causing all 3 blocks to accelerate at the same rate. The tension in the string between blocks B and C is caused by the horizontal force, F, which is pulling all 3 blocks.

The string between blocks B and C is pulling blocks B and A. The 3.00 N force is causing the mass of B + A to accelerate.

F = m * a
3.00 = 0.800 * a
Acceleration = 3.00 ÷ 0.800 = 3.75 m/s^2
This is the acceleration of all 3 blocks!

The horizontal force, F = (mass of 3 blocks) * 3.75
F = 1.2 * 3.75 = 4.5 N

T(AB) is only pulling block A.
T(AB) = 0.4 * 3.75 = 1.5 N

The tension force increase as more mass is being accelerated

genius_generous answered 2 years ago
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