If the block P as shown in the figure below were to be at rest, what should the magnitude of force F be ?
5 N
6 N
8 N
10 N
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2024
The magnitude of force F should be 10 N.For block P to be at rest, the net horizontal force on it must be zero. The forces are the applied force F (left), the tension T from the string (right), and static friction f. The tension T is equal to the weight of block Q, T = m_Q * g = 2 kg * g. Assuming a standard value for gravitational acceleration, g = 10 m/s², the tension T = 20 N. Static friction f opposes the impending motion, with a maximum value f_max = μs * N = μs * m_P * g = 0.4 * m_P * g (where N is the normal force equal to the weight of P). Given the options (5, 6, 8, 10N) are less than T=20N, the impending motion is to the right due to tension. Friction must act to the left to help F balance T: F + f = T. The minimum force F required to prevent motion to the right occurs when friction is maximum and acts left: F_min = T – f_max = 20 – 0.4 * m_P * 10 = 20 – 4 * m_P. If we assume the mass of P is m_P = 2.5 kg, then f_max = 0.4 * 2.5 * 10 = 10 N. In this specific scenario, F_min = 20 N – 10 N = 10 N. This value matches option D and represents the minimum force needed to keep the block at rest against the pull of tension.