sir lancelot, who weighs 800 n, is assaulting a castle by climbing a uniform ladder that is 5.0 m long and weighs 180 n. the bottom of the ladder rests on a ledge and leans across the moat in equilibrium against a frictionless, vertical castle wall. the ladder makes an angle of 53.1 deg with the horizontal. lancelot pauses one third of the way up the ladder. (a) find the normal force and friction forces on the base of the ladder. (b) find the minimum coefficient of static friction needed to prevent slipping at the base. (c) find the magnitude and direction of the contact force on the base of the ladder. (that is, the vector sum of the forces at the base.)

Answers

Answer 1

(a) The normal force is 267.7 N and frictional force is 980 N.

(b) The minimum coefficient of static friction is 0.27, which is needed to prevent slipping at the base.

(c) The direction of the contact force is 1016 N.

∈(t) = 0

N₂ ( L sinθ ) = 800 ( [tex]\frac{L}{3}[/tex] cosθ )

N₂ = 267.7 N

(a) Normal force F(d) = N₂ = 267.7 N

Frictional force N = 800 + 180

                         N = 980 N

(b) The minimum coefficient of static friction,

F(d) = μ N₁

μ = F / N₁

μ = 267.7 / 980

μ = 0.27

(c) The direction of the contact force,

R = [tex]\sqrt{(980)^{2} + (267.7)^{2} }[/tex]

R = 1016 N

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Related Questions

a vertical spring scale can measure weights up to 210 n . the scale extends by an amount of 13.5 cm from its equilibrium position at 0 n to the 210 n mark. a fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.15 hz . part a ignoring the mass of the spring, what is the mass m of the fish?

Answers

sheesh sheesh                                                    

Explanation:

Maddy tripped and accidentally threw her phone off the hollywood cliff at 5m/s. It landed 20 away from the base of the cliff. how high is the cliff then?

Answers

The height of the cliff, given that her phone landed 20 m away when she accidentally threw her phone with a velocity of 5 m/s is 78.4 m

How do I determine the height of the cliff?

To determine the height of the cliff, we must first obtain the time taken for her phone to reach the ground. This is obtained as follow:

Horizontal velocity (u) = 5 m/sHorizontal distance (s) = 20 mTime taken (t) =?

s = ut

20 = 5 × t

Divide both sides by 5

t = 20 / 5

t = 4 s

Finally, we shall determine the height of the cliff. Details below:

Time taken (t) = 4 sAcceleration due to gravity (g) = 9.8 m/s²Height (h) = ?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the height of the cliff is 78.4 m

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when you looked at solar activity on the heliophysics event registry website, on which date did you find activity on the sun?

Answers

You did look at solar activity on the heliophysics event registry website, and we found activity on the sun on the specified day.

What three sorts of solar occurrences are there?

Any natural occurrence in the upper atmosphere of the sun is referred to as a solar phenomenon. Solar flares, prominences and filaments, and coronal mass ejections are three different categories of solar phenomena.

How can I follow the cycle of solar activity the best?

The simplest way we can investigate how solar activity changes over time is to survey sunspots, which is also how we track the solar cycle. Sunspots are associated with the Sun's natural 11-year cycle, during which the planet's surface changes from calm to turbulent.

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The diagram below represents a wave.
What is the speed of the wave if its wavelength is 3.0 m?

Answers

The speed of the wave is determined as 18 m/s.

option D is the correct answer.

What is the speed of the wave?

The speed of the wave is the rate of change of wave's displacement with time.

The speed of the wave is calculated by applying the formula relating speed, wavelength and frequency of the wave as shown below.

v = fλ

where;

v is the speed of the wavef is the frequency of the waveλ is the wavelength of the wave

The frequency of the wave is calculated as follows;

f = 1 / T

where;

T is the period of the wave

The period of a wave is the time taken for the wave to complete one cycle.

In the picture given, 3 cycles of the wave = 0.5 s

1 cycle of the wave = ?

= 0.5 s / 3

= 0.16667 s

f = 1 / T

f = 1 / 0.16667

f = 6 Hz

The speed of the wave is calculated as follows;

v = fλ

v = 6 Hz  x  3 m

v = 18 m/s

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what is the effective spring constant k of the two-spring system? express the effective spring constant in terms of k 1 and k 2 .

Answers

The Effective Spring Constant is [tex](\frac{k1k2}{k1 + k2} )[/tex] for a system of 2 springs with spring constant k1 and k2.

What is Spring Constant ?

The term "spring constant" refers to the proportional constant k. It gauges the stiffness of the spring. A spring exerts a force F = -kx in the direction of its equilibrium position when it is stretched or compressed to a length that differs by an amount x from its equilibrium length.

Spring constants are available in plural (physics) a quality of a spring that is determined by the force acting on the spring to the displacement it produces.

K stands for the proportionality constant, commonly referred to as the "spring constant." In layman's words, stiffness and strength are shown by the k variable in Hooke's law (F = -kx). An item requires more force to be stretched to a specific length the greater the value of k.

Let the force actiong be F N

then , if k1 is the spring constant of spring 1

F = x1 * k1

or, x1 = F/k1.

Similarly for spring 2

x2 = F/k2

Now x1 + x2 = F (1/k1 + 1/k2)

or effective displacement

X= [tex]F(\frac{1}{k1} + \frac{1}{k2} )[/tex]

or, F = [tex]X * (\frac{k1k2}{k1 + k2} )[/tex]

So the Effective Spring Constant is [tex](\frac{k1k2}{k1 + k2} )[/tex].

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1. a 1.4 x 103 kg car is westbound at a velocity of 37.0 km/h when it collides with a 2.0 x 103 kg truck northbound at a velocity of 35 km/h. if these two vehicles lock together upon collision, what is the initial velocity of the vehicles after collision? (7.2 m/s 37o w of n) 2. a 6.2 kg object heading north at 3.0 m/s collides with an 8.0 kg object heading west at 3.5 m/s. if these two masses stick together upon collision, what is their velocity after collision? (2.4 m/s 56o w of n) 3. a 4.0 x 104 n truck moving west at a velocity of 8.0 m/s collides with a 3.0x104 n truck heading south at a velocity of 5.0 m/s. if these two vehicles lock together upon impact, what is their velocity?(5.0 m/s 25o s of w)

Answers

The velocity of vehicles after the collision is 15.82m/sec if car mass is 1.4 × 10³kg and truck mass is 2×10³kg

We know that we need to conserve the momentum of watermelon in a certain direction. Also,we know that initial momentum of watermelon is zero,it means that according to law of conservation of momentum final momentum should be zero.

We know that momentum =mass × velocity

So,initial momentum of car is =1.4 × 10³kg × 37km/hr

Similarly,,initial momentum of truck=2×10³kg × 35km/hr

Now,it is given that both vehicles lock together,so total mass of both vehicles is =1.4 × 10³kg +  2×10³kg=3.4 ×10³kg

Assume final velocity of both vehicle is v

So,following the law of conservation,we get

=>1.4 × 10³kg × 37 + 2×10³kg × 35 = 3.4 ×10³ ×v

=>51.8 ×10³ + 2×10³=3.4×10³ ×v

=>53.8 ×10³ kg-m/sec = 3.4×10³ ×v

=>v = 53.8/3.4

=>v=15.82m/sec

Hence,final velocity of both vehicle after collision is 15.82m/sec.

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5. if the star sirius emits 23 times more energy than the sun, why does the sun appear brighter in the sky? include the terms luminosity and apparent brightness in your response.

Answers

The star Sirius is indeed brighter but our Sun appears to be brighter because it is closer to us.

The intrinsic luminosity of Sirius is exactly 25.4 times more than the sun but if we compare the distances of these stars, Sirius is 8.6 light years away from us and the Sun is just 8 light minutes away from us. This marks the obvious reason why our Sun appears more bright.

Even mathematically, it can be justified as the apparent brightness of any luminous object depends on the inverse square of the distance of that luminous object. Therefore the sun's light energy dominates our sky and we can't see any traces of Sirius from the earth.

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