In: Physics
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 . What is the magnitude F of the force? What is the tension TAB in the string between block A and block B? Express your answers numerically in newtons
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 Tbc= 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 =( 0.400+0.400) * a
Acceleration = 3.00 ÷*( 0.800) = 3.75 m/s^2
This is the acceleration of all 3 blocks is 3.75
The horizontal force, F =MASS *ACCELERATION
F = (mass of 3 blocks = Ma+Mb+Mc) * 3.75
F = 1.2 * 3.75 = 4.5 N
force F =
4.5N
T(AB) is only pulling block A.
T(AB) = m*f
tension ab = 0.4 * 3.75 = 1.5 N
Tab = 1.5
N
so the tension and the forces are calculated