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dybincka [34]
3 years ago
15

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

on at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
Physics
1 answer:
Colt1911 [192]3 years ago
5 0

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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Andrew [12]

Answer:

1.48m

Explanation:

Info given:

mass: 0.772kg

PE = 11.2J

g = 9.81m/s^2

Using KEbottom = PEtop

PE = mgh

Plug in and solve for h (height)

11.2J = mgh

h =11.2J / mg

h = 11.2J / (0.772kg)(9.81m/s^2)

h = 1.48m

4 0
3 years ago
Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)
Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

a' = 3.569

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

v_{f} = v_{o} + a \cdot t

Where:

v_{o} - Initial speed, measured in meters per second.

v_{f} - Final speed, measured in meter per second.

a - Acceleration, measured in \frac{m}{s^{2}}.

t - Time, measured in seconds.

t = \frac{v_{f}-v_{o}}{a}

t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)}{35\,\frac{m}{s^{2}} }

t = 0.095\,s

Lastly, the severity index is now determined:

SI = \frac{a'^{5}}{2\cdot t}

SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}

SI = 3047.749

b) The initial and final speed of the injured are 1.944\,\frac{m}{s} and 5.278\,\frac{m}{s}, respectively. The travelled distance can be determined from this equation of motion:

v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s

Where \Delta s is the travelled distance, measured in meters.

\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}

\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}

\Delta s = 0.345\,m.

8 0
3 years ago
A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. App
monitta

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

4 0
3 years ago
The critical angle for total internal reflection occurs when:___________.
Anastaziya [24]

Answer:

b) the refracted ray has an angle of 90 degrees

Explanation:

The angle of incidence is measured with respect to the normal separation of the media. The critical angle for total internal reflection occurs is:

\theta=arcsin(\frac{n_2}{n_1})

Here n_2 and n_1 are the refractive index of the mediums. This equation is an application of Snell's law, for the case where  the refracted ray has an angle of 90^\circ.

5 0
3 years ago
I need help in my physics class and show me how it’s done
Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846

So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

\alpha = -47.28 ° +360° = 313 °

\alpha = 313 °

Option 4.

4 0
3 years ago
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