Two masses, m_1 and m_2, are connected by a mass-less cord passing over an ideal
ID: 1604133 • Letter: T
Question
Two masses, m_1 and m_2, are connected by a mass-less cord passing over an ideal pulley as shown in the figure below. Assume that m_1 is moving on a frictionless surface. a. Draw a free body diagram for each mass showing all the forces and accelerations. b. Write Newton's equations for each mass. c. Show that the formula for the acceleration of the system in terms of m_1, m_2 and g is give by, a = (m_1/m + m_2)g d. Find the tension on the string. e. Explain why the equations for acceleration and tension make sense when either m_1 = 0 or m_2 = 0. f. If the masses are in equilibrium, what can you say about their motion? g. How would friction between m_1 and the table would change your free body diagrams, equations and answers? Two crates connected by a rope lie on a horizontal surface as shown in the figure below. Crate A has mass m_A and crate B has mass m_B. The coefficient kinetic friction between each crate and the surface is the of same and equal to mu_k. The crates are pulled to the right at by a horizontal force F.Explanation / Answer
2)
b) Let 'a' be the acceleration of the system and 'T' be the tension in the string
Newton's equation for m1: T = m1*a .........(1)
Newton's equation for m2: m2g - T = m2*a ..........(2)
Substituting (1) in (2),
m2g - m1*a = m2*a
=> m2g = (m1 + m2)*a
=> a = m2g / (m1 + m2)
c) T = m1*a
=> T = m1*m2*g / (m1 + m2)
d) If m2 =0 => a = 0 (m1 will be stationary)
If m1 =0 => a = g (m2 will free fall)
If either m1 or m2 is 0, T=0 (string will not be taut)
e) If the masses are in equilibrium, acceleration is 0
T = m2g
f) If 'f' is the friction between m1 and the surface,
Newton's equation for m1: T - f = m1*a
Newton's equation for m2: m2g - T = m2*a