What's the thrust force of 20,000N

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

3 years ago

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in

Kazeer [188]

Answer:

Explanation:

it equals16 2211

8 0

3 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

$u_y = u sin \theta$

where

u = 21.5 m/s is the initial speed

$\theta=33.5^{\circ}$ is the angle

Substituting,

$u_y = (21.5) sin 33.5^{\circ} =11.9 m/s$

The vertical position of the ball at time t is given by

$y = h + u_y t + \frac{1}{2}gt^2$

where

h = 13.5 m is the initial heigth

$g = -9.8 m/s^2$ is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

$0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2$

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

$u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s$

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

$d= u_x t = (17.9)(3.27)=58.5 m$

3 0

3 years ago

On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2 )/6 = 1.63 m/s2 .

MAVERICK [17]

The pendulum would differ from 300 inches

8 0

3 years ago

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w

mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg = V²μ

g = V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

7 0

3 years ago