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Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
Answer:
a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.
b) The ball is in the air during approximately 2.890 seconds.
c) The horizontal distance covered by the ball is 32.695 meters.
d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.
e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.
Explanation:
a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:
(1)
Where:
,
- Initial and final vertical position, measured in meters.
- Initial speed, measured in meters per second.
- Launch angle, measured in sexagesimal degrees.
- Gravitational acceleration, measured in meters per square second.
- Time, measured in seconds.
If we know that ,
,
,
and
, then the time taken by the ball is:
(2)
This second order polynomial can be solved by Quadratic Formula:
and
Only the first root offers a solution that is physically reasonable. That is, .
The vertical velocity of the ball is calculated by this expression:
(3)
Where:
,
- Initial and final vertical velocity, measured in meters per second.
If we know that ,
,
and
, then the final vertical velocity is:
The y-component of velocity just before the ball hits the ground is -14.860 meters per second.
b) From a) we understand that ball is in the air during approximately 2.890 seconds.
c) The horizontal distance covered by the ball () is determined by the following expression:
(4)
If we know that ,
and
, then the distance covered by the ball is:
The horizontal distance covered by the ball is 32.695 meters.
d) The magnitude of the velocity of the ball just before hitting the ground (), measured in meters per second, is determined by the following Pythagorean identity:
(5)
If we know that ,
and
, then the magnitude of the velocity of the ball is:
.
The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.
e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:
If we know that ,
and
, the angle of the total velocity of the ball just before hitting the ground is:
The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.