1. Find the tension force T_{1}, T_{2}, and T_{3}. Ignore cord’s mass.

Solution

(a) Free-body diagram for object (b) Free-body diagram for cord

Apply the Newton’s first law on the object :

ΣF_{y} = 0

T_{1} – w = 0

T_{1} = w = m g

T_{1} = (5 kg)(9.8 m/s^{2})

T_{1 }= 49 kg m/s^{2}

**T**_{1}** = 49 N**

Apply Newton’s first law on the cord :

∑F_{x} = 0

T_{3x} – T _{2x }= 0

T_{3 }cos 30^{o} – T_{2} cos 40^{o }= 0

0.87 T_{3 }– 0.77 T_{2 }= 0

0.87 T_{3} = 0.77 T_{2}

T_{2 }= 0.87 T_{3 }/ 0.77 = 1.1 T_{3 }———- Equation 1

[irp]

——

∑F_{y} = 0

T_{3y} + T_{2y} – T_{1y} = 0

T_{3 }sin 30^{o} + T_{2 }sin 40^{o} – T_{1} = 0

0.5 T_{3 }+ 0.64 T_{2 }– 49 N = 0 ———- Equation 2

Substituting T_{2 }in the equation 2 into the equation 2 :

0.5 T_{3 }+ 0.64 (1.1 T_{3}) – 49 N = 0

0.5 T_{3} + 0.70 T_{3 }– 49 = 0

1.2 T_{3} – 49 = 0

1.2 T_{3 }= 49

T_{3} = 49 / 1.2

**T**_{3}** = 41 N**

———

T_{2} = 1.1 T_{3}

T_{2} = (1.1)(40.8 N)

**T**_{2}** = 45 N**

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