Consider the same three blocks (m_1 = 2.60 kg, m_2 = 5.46 kg, and m_3 = 5.20 kg)
ID: 1413925 • Letter: C
Question
Consider the same three blocks (m_1 = 2.60 kg, m_2 = 5.46 kg, and m_3 = 5.20 kg), strings, and tabic s>> in the previous scenario. Now the positions of the blocks are switched so the m_3 is in the middle (in contact with the table) and m_2 is on the right (hanging). The coefficients of friction between the middle block and the table are the same values as before: = 0.350 and fx_s = 0.500, respectively. mu The blocks are released from rest simultaneously. Determine the magnitude of the acceleration of the left block (m_1). Determine the tension in the left string. Determine the tension in the right string. Determine the magnitude of the frictional force exerted on the middle block by the table.Explanation / Answer
M1 = 2.6
M2= 5.469
m3=5.2
us=0.5
uk = 0.35
m2 >m1
T2 -T1 = (5.469-2.6)*9.8 = 28.1162 N >( 0.5*9.8*5.2)
so M3 will move
towards m2 direction
m2 * g - T2 = m2*a
T2-T1 - 0.35*9.8*5.2 = 5.2*a
T1 - m1*g = m1*a
adding all we get
a = 0.774 m/s^2
T1 =tension in left = 27.49
T2 = tension in right = 49.36 N
Friction = 0.35*9.8*5.2 = 17.836 N ( opposite to motion direction)