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Llana [10]
2 years ago
11

A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on the free end of the rope f

or maintaining equilibrium that is required to just support a 5 kN weight. The coefficient of friction between the rope and the cylinder is 0.2
Physics
1 answer:
Taya2010 [7]2 years ago
8 0

The range of force exerted at the end of the rope is 285.7 N to 1,000 N.

<h3>Net horizontal force of the cylinder</h3>

The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.

∑F = 0

F - μFn = 0

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force increases as the number of turns of the rope increases.

minimum force = total force/number of turns of rope

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Learn more about Newton's second law of motion here: brainly.com/question/3999427

You might be interested in
A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m
Nina [5.8K]

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

Extension of wire \delta l=0.1\ mm

Extension in a wire is given by

\Rightarrow \delta l=\dfrac{FL}{AE}

where, E=\text{Youngs modulus}

\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)

for same force, length and material

\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm

5 0
3 years ago
PLEASE PROVIDE AN EXPLANATION.<br><br> THANKS!!!
ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(c) 1250 N

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

(c) The tension is:

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))

y = 0.0365 m

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