PROBLEM 3: Three blocks, connected together by mass-less strings as shown in the

Question

PROBLEM 3: Three blocks, connected together by mass-less strings as shown in the picture below, are being pulled along by a horizontal force F. The tension magnitude in the string between blocks B and C is 3.5 N. The tension magnitude in the string between blocks A and B is unknown. Let each block have the following mass: 205 grams for block A, 415 grams for block B, and 552 grams for block C. The coefficient of kinetic friction for each block is 0.17. lgnore air resistance. (a) Calculate is the magnitude of the acceleration of block B

Explanation / Answer

Here ,

T = 3.5 N

mA = 0.205 Kg

mB = 0.415 Kg

mC = 0.552 Kg

a)

Here , for blocks A and B

using second equation of motion

T - uk * (mA + mB) *g = (mA + mB) * a

3.5 - 0.17 * (0.205 + 0.415) * 9.8 = (0.205 + 0.415) * a

a = 4 m/s^2

the acceleation of block B is 4 m/s^2

b)

let the tension between A and B is TAB

TAB - 0.17 * 0.205 * 9.8 = 0.205 * 4

TAB = 1.16 N

the tension between blocks A and B is 1.16 N

c)

Now , for block C

F - T - u * mC * g = mC*a

F - 3.5 - 0.17 * 0.552 * 9.8 = 0.552 * 4

solving for F

F = 6.63 N

the magnitude of force F is 6.63 N

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