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3 years ago

"A window washer pulls herself upward using the bucket-pulley apparatus. The mass of the person plus the bucket is 65 kg.

Physics

2 answers:

FromTheMoon [43]3 years ago

8 0

Part A.

If there is a constant speed, therefore the net force is zero.

Forces acting are the:

1. Weight

2. Upward force exerted by the rope which is opposite in direction but the magnitude is to the force which is acting downwards exerted on the rope.

∑Force = pulling force - weight = 0

Weight = mg

Hence,

F (pulling force) = mg

Substituting the values,

F (pulling force) = 65 × 9.81

F (pulling force) = 637.65 Newton

____________________________________________________________

Part B.

F = ma

10% increase in the force = 1.10 ₓ 637.65

10% increase in the force = 701.415

Therefore,

701.415 = 65 ₓ a

a = 701.45/65

a = 10.8 m/s²

Aleksandr [31]3 years ago

7 0

Consider the free body diagrams as shown in the file attached.

Let the acc of system is a. <span>
For the washer T+N-m1g=m1a ............................(1)
For the bucket T-N-m2g=m2a ............................(2)
</span>
(A)
a=0 Eqns (1) and (2) become T+N-m1g=0 T-N-m2g=0 Adding and rearranging, T=(m1+m2)/2g T=325N <span>
(B)
</span> We are specified the value of T here as 325+32.5=357.5N eqns (1)and (2) become 357.5+N-m1g=m1a 357.5-N-m2g=m2a Adding, 715-(m1+m2)g=(m1+m2)a

Solve for 'a'

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

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The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

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The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

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$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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