Three blocks, of masses m1=2.0, m2=4.0 and m3=6.0 kg are connected by strings on a frictionless inclined plane of 60°, as shown J the figure. A force F=120 N is applied upward along the incline to the uppermost block, causing an upward movement of the blocks, if connecting cords are light. The values of tensions T1 and in the cords are

Question

Three blocks, of masses m1=2.0, m2=4.0 and m3=6.0 kg are connected by strings on a frictionless inclined plane of 60°, as shown J the figure. A force F=120 N is applied upward along the incline to the uppermost block, causing an upward movement of the blocks, if connecting cords are light. The values of tensions T1 and in the cords are (A) T1=20 N, T2=60 N  (B)  T1=60 N, T2=60 N  (C) T1=30 N, T2=50 N  (D) T1=20 N, T2=100 N

Tardigrade Admin · Accepted Answer

T2=((m1+m2) F/(m1+m2+m3)) =((2+4) × 120/(2+4+6))=(6 × 120/12) T2=60 N T1=(m1 F/(m1+m2+m3))=(2 × 120/12) = 20 N

$T_{1} = 20 N, T_{2} = 60 N$

$T_{1} = 60 N, T_{2} = 60 N$

$T_{1} = 30 N, T_{2} = 50 N$

$T_{1} = 20 N, T_{2} = 100 N$

$T_{2} = \frac{( m _{1} + m _{2} ) F}{( m _{1} + m _{2} + m _{3} )}$
$= \frac{( 2 + 4 ) \times 120}{( 2 + 4 + 6 )} = \frac{6 \times 120}{12} T_{2} = 60 N$
$T_{1} = \frac{m _{1} F}{( m _{1} + m _{2} + m _{3} )} = \frac{2 \times 120}{12}$
$= 20 N$

T1​=20N,T2​=60N

T1​=60N,T2​=60N

T1​=30N,T2​=50N

T1​=20N,T2​=100N

T2​=(m1​+m2​+m3​)(m1​+m2​)F​ =(2+4+6)(2+4)×120​=126×120​T2​=60N T1​=(m1​+m2​+m3​)m1​F​=122×120​ =20N

$T_{1} = 20 N, T_{2} = 60 N$

$T_{1} = 60 N, T_{2} = 60 N$

$T_{1} = 30 N, T_{2} = 50 N$

$T_{1} = 20 N, T_{2} = 100 N$

$T_{2} = \frac{( m _{1} + m _{2} ) F}{( m _{1} + m _{2} + m _{3} )}$
$= \frac{( 2 + 4 ) \times 120}{( 2 + 4 + 6 )} = \frac{6 \times 120}{12} T_{2} = 60 N$
$T_{1} = \frac{m _{1} F}{( m _{1} + m _{2} + m _{3} )} = \frac{2 \times 120}{12}$
$= 20 N$