A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
2 answers:
The hi.nge will be subjected to a force of 261.12N from the beam.
We must understand the tension in order to choose the solution.
<h3>How can the amount of force the beam applies on the height be determined?</h3>- Let's use the provided information to create the system's free body diagram.
- We need to calculate the force the beam is exerting on the height using the diagram.
- For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.
- We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.
- Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.
- As a result, the normal reaction that the beam has on the height becomes, with a vertical component,
- As a result, the force the beam applies on the height will be of the order of,
Thus, we can infer that the force the beam applies to the height is 261.12N in size.
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The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.
- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.
- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.
- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,
- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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Answer:
54.9 m/s at 44.9 degrees
Explanation:
If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.
Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s
Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s
SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.
Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use
Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2
We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m
If we take the first derivative of the equation of position, we get the equation for speed
V(t) = Vy0 + a * t
We know that being t2 the moment the ball goes over the wall
V(t2) = -25 m/s
Y(t2) = 45.4 m
So:
45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2
-25 = Vy0 + a * t2
Then:
Vy0 = -25 - a * t2
So:
45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2
0 = -44.4 - 25 * t2 - 1/2 * a * t2^2
a = -9.81 m/s^2
0 = -44.4 - 25 * t2 + 4.9 * t2^2
Solving this quadratic equation we get:
t1 = -1.39 s
t2 = 6.5 s
Since we are looking for a positive value we disregard t1.
Now we can obtain Vy0:
Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s
Since horizontal speed is constant Vx0 = 38.9 m/s
By Pythagoras theorem we obtain the value of the initial speed:
The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90
a = atan(Vy0/Vx0) = 44.9 degrees
Answer:
think its b i dunno
Explanation:
