1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
BaLLatris [955]
3 years ago
6

1) A uniform wooden beam, with mass of 120 and length L = 4 m, is supported as illustrated in the figure. If the static friction

coefficients at points A and B ua= 0.3 and ub=0.1 determine the value of the angle so that the system is in equilibrium if (a) =70 and (b)=90
2) In the situation illustrated below, the blocks of mass Ma=1.0 kg and Mb=2.5 kg are connected by a rope and pulley (of negligible mass and friction) and are on inclined planes with =30 and =50 If the coefficients of kinetic friction between the blocks and the surfaces are ua=0.30 and ub=0.15 determine (a) the tension on the rope and (b) the acceleration of the blocks


3) In the figure below, a ball is thrown from point A on top of an inclined plane of = 57°, with an initial velocity V0 and an angle . The ball takes 1.25 seconds to reach the maximum height of its trajectory (point B) and at this moment it is at a horizontal distance D = 3.7 m from the launch point. For this situation, calculate (a) the launch angle; (b) the initial velocity V0; (c) the total flight time between points A and C; (d) the velocity of the ball at the instant it reaches point C; and (e) the distance D between the launch point and the impact point. Tip: use the relationship
sen(a)/cos(a)=tg(a)



4) In the figure below, a car leaves from rest at = 0 with an acceleration given by at = (0.7t) m/s towards a radius curve R = 90 m. Considering that it is initially at a distance d = 140 from the curve, calculate the module of your total acceleration in (a) t = 10.5 s and (b) t = 12.2 s. If the car started from rest with an At = 3 m/s acceleration, (c) what would be the module of its total acceleration after having
displaced of 187 m?

Physics
1 answer:
Kobotan [32]3 years ago
5 0

Answer:

1(a) 55.0°

1(b) 58.3°

2(a) 10.2 N

2(b) 2.61 m/s²

3(a) 76.7°

3(b) 12.8 m/s

3(c) 3.41 s

3(d) 21.8 m/s

3(e) 18.5 m

4(a) 7.35 m/s²

4(b) 31.3 m/s²

4(c) 12.8 m/s²

Explanation:

1) Draw a free body diagram on the beam.  There are five forces:

Weight force mg pulling down at the center of the beam,

Normal force Na pushing up at point A,

Friction force Na μa pushing left at point A,

Normal force Nb pushing perpendicular to the incline at point B,

Friction force Nb μb pushing up the incline at point B.

There are 3 unknown variables: Na, Nb, and θ.  So we're going to need 3 equations.

Sum of forces in the x direction:

∑F = ma

-Na μa + Nb sin φ − Nb μb cos φ = 0

Nb (sin φ − μb cos φ) = Na μa

Nb / Na = μa / (sin φ − μb cos φ)

Sum of forces in the y direction:

∑F = ma

Na + Nb cos φ + Nb μb sin φ − mg = 0

Na = mg − Nb (cos φ + μb sin φ)

Sum of torques about point B:

∑τ = Iα

-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0

mg (L/2) cos θ = Na L cos θ − Na μa L sin θ

mg cos θ = 2 Na cos θ − 2 Na μa sin θ

mg = 2 Na − 2 Na μa tan θ

Substitute:

Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)

1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)

2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)

Substitute again:

2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)

tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)

a) If φ = 70°, then θ = 55.0°.

b) If φ = 90°, then θ = 58.3°.

2) Draw a free body diagram of each mass.  For each mass, there are four forces.  For mass A:

Weight force Ma g pulling down,

Normal force Na pushing perpendicular to the incline,

Friction force Na μa pushing parallel down the incline,

Tension force T pulling parallel up the incline.

For mass B:

Weight force Mb g pulling down,

Normal force Nb pushing perpendicular to the incline,

Friction force Nb μb pushing parallel up the incline,

Tension force T pulling up the incline.

There are four unknown variables: Na, Nb, T, and a.  So we'll need four equations.

Sum of forces on A in the perpendicular direction:

∑F = ma

Na − Ma g cos θ = 0

Na = Ma g cos θ

Sum of forces on A up the incline:

∑F = ma

T − Na μa − Ma g sin θ = Ma a

T − Ma g cos θ μa − Ma g sin θ = Ma a

Sum of forces on B in the perpendicular direction:

∑F = ma

Nb − Mb g cos φ = 0

Nb = Mb g cos φ

Sum of forces on B down the incline:

∑F = ma

-T − Nb μb + Mb g sin φ = Mb a

-T − Mb g cos φ μb + Mb g sin φ = Mb a

Add together to eliminate T:

-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a

g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a

a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)

a = 2.61 m/s²

Plug into either equation to find T.

T = 10.2 N

3i) Given:

Δx = 3.7 m

vᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

t = 1.25 s

Find: v₀ₓ, v₀ᵧ

Δx = v₀ₓ t + ½ aₓ t²

3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²

v₀ₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ

v₀ᵧ = 12.5 m/s

a) tan θ = v₀ᵧ / v₀ₓ

θ = 76.7°

b) v₀² = v₀ₓ² + v₀ᵧ²

v₀ = 12.8 m/s

3ii) Given:

Δx = D cos 57°

Δy = -D sin 57°

v₀ₓ = 2.96 m/s

v₀ᵧ = 12.5 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

c) Find t

Δx = v₀ₓ t + ½ aₓ t²

D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²

D cos 57° = 2.96t

Δy = v₀ᵧ t + ½ aᵧ t²

-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²

-D sin 57° = 12.5t − 5t²

Divide:

-tan 57° = (12.5t − 5t²) / 2.96t

-4.558t = 12.5t − 5t²

0 = 17.058t  − 5t²

t = 3.41 s

d) Find v

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (3.41 s) + 2.96 m/s

vₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s

vᵧ = -21.6 m/s

v² = vₓ² + vᵧ²

v = 21.8 m/s

e) Find D.

D cos 57° = 2.96t

D = 18.5 m

4) Given:

R = 90 m

d = 140 m

v₀ = 0 m/s

at = 0.7t m/s²

The distance to the opposite side of the curve is:

140 m + (90 m) (π/2) = 281 m

a) Find Δx and v if t = 10.5 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (10.5)²

vt = 38.6 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (10.5)³

Δx = 135 m

The car has not yet reached the curve, so the acceleration is purely tangential.

at = 0.7 (10.5)

at = 7.35 m/s²

b) Find Δx and v if t = 12.2 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (12.2)²

vt = 52.1 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (12.2)³

Δx = 212 m

The car is in the curve, so it has both tangential and centripetal accelerations.

at = 0.7 (12.2)

at = 8.54 m/s²

ac = v² / r

ac = (52.1 m/s)² / (90 m)

ac = 30.2 m/s²

a² = at² + ac²

a = 31.3 m/s²

c) Given:

Δx = 187 m

v₀ = 0 m/s

at = 3 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (3 m/s²) (187 m)

v = 33.5 m/s

ac = v² / r

ac = (33.5 m/s)² / 90 m

ac = 12.5 m/s²

a² = at² + ac²

a = 12.8 m/s²

You might be interested in
Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,
aev [14]
9.8 ms^-2 is acceleration
4 0
3 years ago
Speed is a component of skill related fitness what does speed enable you to do.
Mrrafil [7]

I'd say move faster, unless it's asking something else.

7 0
3 years ago
A 1.5kw vacuum cleaner is used for half an hour. Calculate its energy usage.​
juin [17]

Answer:

its energy usage is 0.05

Explanation:

1.5kw/30=0.05

8 0
2 years ago
If τ=r×F then F.τ is equal
Mademuasel [1]
Yes that is correct.
4 0
3 years ago
Help with the two questions above? Correct answers?
LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz

The wavelength is then

\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m

The third choice "0.80m; 431Hz" is correct

7 0
3 years ago
Other questions:
  • If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.
    12·1 answer
  • Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate of a parallel plate capacitor. What is th
    11·1 answer
  • Blocks A (5.00 kg) and B (10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is mov
    12·1 answer
  • The blue mass is currently 4 meters away from the red mass.
    13·1 answer
  • A spring attached to a mass is at rest in the initial
    13·2 answers
  • Explain two ways you could reduce the friction between two surfaces
    13·1 answer
  • Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one
    10·1 answer
  • A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat
    10·1 answer
  • A wire of radius 0.8 cm carries a current of 106 A that is uniformly distributed over its cross-sectional area. Find the magneti
    7·1 answer
  • Deduced hydrochloric acid is a strong acid ​
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!