Question

4. In the arrangement shown in figure what should be the mass of
block A, so that the system remains at rest? Neglect friction and
mass of strings.

Fig. 8.75​

Answer 1

Answer:

3 kg

Explanation:

I'll call blocks B1 B2 B3 (from left to right). mg of B1 is 20N (2 kg * 10 m/s^2), we split mg of B2 into mg*cos(30) and mg*sin(30) (from similar triangles). B2 has a tendency to go down hill and the force that is making it do it is mg*sin(30). In order for the system to remain at rest, B1 + B2 = B3. B1 = 20 n, B2 = 2 kg * 10 m/s^2 * 1/2 (1/2 = sin(30)) = 10 N. 20 N + 10 N = 30 N. That means B3 is 30 N, from which we get 3 kg.