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3 years ago

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A block with mass m1m1m_1 is placed on an inclined plane with slope angle ααalpha and is connected to a second hanging block tha

t has mass m2m2m_2 by a cord passing over a small, frictionless pulley. The coefficient of static friction is μsμsmu_s and the coefficient of kinetic friction is μkμkmu_k. Find the smallest value of m2 when the blocks will remain at rest if they are released from rest.

1 answer:

vredina [299]3 years ago

6 0

Answer:

Explanation:

m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .

For equilibrium of m₁

m₁ g sinα= T + f where f is force of friction

m₁ g sinα= T + μsx m₁ g cosα

m₁ g sinα - μs x m₁ g cosα = T

For equilibrium of m₂

T = m₂g

Putting this value in equation above

m₁ g sinα - μs x m₁ g cosα = m₂g

m₂ = m₁ sinα - μs x m₁ cosα

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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3 years ago

A baseball is moving at a speed of 2.2m/s when it strikes the catchers glove. The paddding of the glove is compressed by 24mm be

Andre45 [30]

Average acceleration of the baseball: $-101 m/s^2$

Explanation:

Since the motion of the baseball is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the displacement of the object

For the baseball in this problem, we have:

u = 2.2 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

s = 24 mm = 0.024 m is the displacement of the ball while decelerating

Therefore, we can solve for a to find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-2.2^2}{2(0.024)}=-101 m/s^2$

where the negative sign means the baseball is slowing down.

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2 years ago

The difference between the frequency fff and the frequency ωωomega is that fff is measured in cycles per second or hertz (abbrev

Kisachek [45]

Answer:

Radians

Explanation:

The angular speed is a measure of the rotation speed of a body. It is defined as the angle rotated by a unit of time. Thus, It refers to the angular displacement per unit time and is designated by the Greek letter $\omega$ . Its unit in the International System is radian per second (rad / s).

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3 years ago

A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W

labwork [276]

Answer:

<h2>650W/m²</h2>

Explanation:

Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

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Area (in m²) = 0.01m²

Required

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Hence, the intensity of the sunlight in W/m² is 650W/m²

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3 years ago

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