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Zanzabum
3 years ago
10

A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is

in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)

Physics
2 answers:
lakkis [162]3 years ago
6 0

Answer:

Tensions of:

DA = 28.81 KN

DB = 16.45 KN

DC = 28.07 KN

Explanation:

see attached

xxMikexx [17]3 years ago
3 0

Answer:

Fad = 28.8 kN

Fbd = 16.4 kN

Fcd = 28.1 kN

Explanation:

First, find the length of each cable.

AD = √((2 m)² + (0.5 m)² + (2.5 m)²)

AD = √10.5 m

AD ≈ 3.24 m

BD = √((1.5 m)² + (1 m)² + (2.5 m)²)

BD = √9.5 m

BD ≈ 3.08 m

CD = √((1 m)² + (1 m)² + (2.5 m)²)

CD = √8.25 m

CD ≈ 2.87 m

Next, use similar triangles to find the x, y, and z components of each tension force.

Fadx = 2/3.24 Fad = 0.617 Fad

Fady = 0.5/3.24 Fad = 0.154 Fad

Fadz = 2.5/3.24 Fad = 0.772 Fad

Fbdx = 1.5/3.08 Fbd = 0.487 Fbd

Fbdy = 1/3.08 Fbd = 0.324 Fbd

Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd

Fcdx = 1/2.87 Fcd = 0.348 Fcd

Fcdy = 1/2.87 Fcd = 0.348 Fcd

Fcdz = 2.5/2.87 Fcd = 0.870 Fcd

Now sum the forces in the x, y, and z directions:

∑Fx = ma

-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0

∑Fy = ma

-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0

∑Fz = ma

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.

-0.463 Fad + 0.811 Fbd = 0

0.811 Fbd = 0.463 Fad

Fbd = 0.571 Fad

Substitute into either of the first two equations:

-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0

-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0

-0.339 Fad + 0.348 Fcd = 0

0.348 Fcd = 0.339 Fad

Fcd = 0.975 Fad

Now substituting into the third equation:

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0

60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0

60 kN − 2.083 Fad = 0

Fad = 28.8 kN

Solving for the other two tension forces:

Fbd = 0.571 Fad = 16.4 kN

Fcd = 0.975 Fad = 28.1 kN

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Answer:

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You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of th
solmaris [256]

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

v_x = u cos \theta

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u = 8.3 m/s is the initial velocity of the balloon

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Substituting,

v_x = (8.3)(cos 27^{\circ})=7.4 m/s

b)

The vertical component of the initial velocity of a projectile is given by

u_y = u sin \theta

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Here we have

u = 8.3 m/s

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Substituting,

u_y = (8.3)(sin 27^{\circ})=3.8 m/s

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The vertical component of the velocity of the balloon follows the suvat equation

v_y = u_y - gt

where

v_y is the vertical velocity at time t

u_y = 3.8 m/s is the initial vertical velocity

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The balloon reaches the maximum height when the vertical velocity becomes zero:

v_y = 0

So we get:

0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s

d)

The maximum height of the balloon can be calculated using the suvat equation:

s=u_y t - \frac{1}{2}gt^2

where

u_y = 3.8 m/s is the initial vertical velocity

g=9.8 m/s^2 is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

t=\frac{2u_y}{g}

where

u_y is the initial vertical velocity

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Here we have

u_y = 3.8 m/s

g=9.8 m/s^2

Substituting,

t=\frac{2(3.8)}{9.8}=0.78 s

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

d=v_x t

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v_x is the horizontal velocity

t is the time of flight

Here we have

v_x = 7.4 m/s

t = 0.78 s

Substituting,

d=(7.4)(0.78)=5.77 m

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