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3 years ago

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A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

:
How far will the load move if it starts from rest and accelerates for 20 seconds?

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

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4 years ago

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a). We got N = mg cos θ,

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T = mg sin θ + f

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Putting (ii) in (i), we get

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b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

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$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

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Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

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$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

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3 years ago