Question

Having difficulty on this problem. Would very much appreciate it if someone could offer some assistance. Thank you.

5. In the system shown, block A weighs 10lbs, block B weighs 40lbs and block C weighs 100lbs. Ail pulleys are frictionless and the cable connects block B and C as shown. A force F(t) is applied to block C causing it to accelerate. 1) Calculate the accelerations of block B and C just before block A begins to slide. The coefficent of static friction between A and B is 0.4. The coefficient of kinetic friction between B and the ground is 0.2. 2) Next, F(t) changes such that the acceleration of block C increases just slightly. After this slight increase, the acceleration of block C remains constant. This causes block A to now slide on block B. It is observed that block A slides 10 inches on top of block B in 0.5 seconds. Calculate the acceleration of block A during the sliding. 3) Extra Credit: Calculate the Tension in the cable just before A starts to slide - OR - Calculate the coefficient of kinetic friction (uk) on the surface between block A and B.

Answer 1

Given that,

     A three mass bodies which aare connected each other to their moments.

a)   Calculate the accelerations of the block B and block C

    $\mu$_k for A and B = 0.4              $\mu$_k for B and ground = 0.2

    F = $\mu$_kN

      For A and B the normal force N = mg = 22.64 kg X 10 = 226.4 kg -m /s²

        F = 0.4 X 226.4 = 90.56N

      F = ma

    a = F / m = 90.56 / 22.64 = 4 m / s²

Now for the acceleration of C

      F = $\mu$_k N

    N = mg = 45.3 kg X 10 = 453 kg - m / s²

       F = 0.2 X 453 = 90.6 N

   F = ma

a = F / m

    = 90.6 / 45.3

   = 2 m / s²

2) We know that,

   S = ut + 1/2 at²

given S = 10 inches = 0.25m

      t = 0.5 s      u = 0

0.25 = 0 X 0.5 + 1/2 a X 0.5²

   0.25 = 0.125 a

a = 2 m / s²

c) Calculate the tension in the cable just before A starts to slide.

     F(t) = ?

    F(t) = sum of all the forces

           = Normal force + friction force

          = 90.56 + 90.6

        = 181.16 N