A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 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.)
1 answer:
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
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Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Data Given:
Electric Field between two parallel plates = 628 N/C
Separation = 4.22 cm
a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.
Solution:
First of all:
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=
E
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a =
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a = Equation 1
So,
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S = ut + 1/2
Here, u = 0.
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Put equation 1 into the above equation:
S = 1/2 x ( )
Equation 2
So,
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(D-S) = 1/2 x ( )
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We know that the charge of electron is equal to the charge of proton so,
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= q
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S
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S =
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we already have the equation, we need to put the values in it.
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