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kolezko [41]
4 years ago
13

â/8.37 points scalcet8 12.2.037. ask your teacher my notes question part points submissions used a block-and-tackle pulley hoist

is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 350 n. the ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. find the tension in each rope and the magnitude of each tension. (let t2 and t3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. round all numerical values to two decimal places.)

Physics
1 answer:
AnnyKZ [126]4 years ago
4 0
Refer to the diagram shown below.

The hoist is in static equilibrium supported by tensions in the two ropes.

For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂             (1)

For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350     (2)

Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979

Answer:
T₂ = 225.12 N
T₃ = 275.98 N

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Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle
tangare [24]

Answer:

F=1.159

Explanation:

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Mass of pulley M=1kg

Radius r=12cm

Mass of block A M_a=2.1kg

Mass of block B m_b=4.1kg

Spring constant\mu= 358 J/m2

Generally the equation for Torque is mathematically given by

Since \sumF=ma

At mass A

 T_2-f_3=2.1a

At mass B

 4.8-T_1=4.1a

At  Pulley

 R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}

 R(T_1-T_2)=0.55a

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a

 (T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a

 -f_3+4.1=6.75a

 -f_3=6.75a+4.8

Since From above equation

M_{eff}=6.7kg

Therefore

T=2\pi \sqrt{{\frac{M_{eff}}{k}}

T=2\pi \sqrt{{\frac{6.75}{\mu}}

T=0.862s

Generally the equation for frequency is mathematically given by

F=\frac{1}{T} \\F=\frac{1}{0.862}

F=1.159

3 0
3 years ago
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