A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 25° above the horizontal. If the normal force exerted on the suitcase is 155 N, what is the force F applied to the handle?
and..
A paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced an upward normal force of approximately 261,500 N acting on its torso when it hit the ground. Assume the torso has a mass of 3720 kg.
and i foudn that the magnitude is 60.4857 when it comes to rest
(b) Assuming the torso is in free fall for a distance 1.51 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?
To find the force applied to the handle in the first question, we need to consider the forces acting on the suitcase. Since the suitcase is being pulled with a constant speed (no acceleration), the net force acting on it must be zero.
The forces acting on the suitcase are:
1. Gravitational force (weight) = m * g, where m is the mass of the suitcase (23 kg) and g is the acceleration due to gravity (9.8 m/s^2).
2. Normal force = 155 N, which is perpendicular to the horizontal surface.
Since the suitcase is being pulled at an angle of 25° above the horizontal, we need to resolve the force applied into horizontal and vertical components. Let's call the force applied to the handle F.
The vertical component of the force applied is F * sin(25°), which opposes the gravitational force. So, we have the equation:
F * sin(25°) = m * g
The horizontal component of the force applied is F * cos(25°), which is balanced by the frictional force (assuming the suitcase is on a horizontal surface with no inclination). Since the suitcase is moving at a constant speed, the frictional force must be equal in magnitude and opposite in direction to the horizontal component. Therefore, we have:
F * cos(25°) = frictional force
Since there is no vertical acceleration, the gravitational force and the normal force must cancel each other out. So, we have:
m * g = normal force = 155 N
Now, we can solve these equations simultaneously to find the force applied to the handle, F.
For the second question, we need to calculate the time it takes for the torso to come to rest once it contacts the ground.
To find the time, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the torso comes to rest)
u = initial velocity (unknown)
a = acceleration (unknown)
s = distance (1.51 m)
Initially, the torso is in free fall, so the only force acting on it is the gravitational force. Using the equation:
force = mass * acceleration
We can rearrange it to find the acceleration:
acceleration = force / mass
In this case, the force is the normal force experienced by the torso, which is approximately 261,500 N, and the mass is 3720 kg. So, we have:
acceleration = 261,500 N / 3720 kg = 70.45 m/s^2 (approximately)
From the problem, we know that the torso comes to rest after falling a distance of 1.51 m. So, the final velocity, v, is 0 m/s, and the distance, s, is 1.51 m.
Using the kinematic equation, we can rearrange it to find the initial velocity, u:
u^2 = v^2 - 2as
u^2 = 0^2 - 2 * 70.45 m/s^2 * 1.51 m
u^2 = - 213.0845 m^2/s^2 (approximately)
Note that the negative value is expected, as the torso is decelerating (opposite direction to initial motion).
Now, we can substitute the values and solve for u:
u = sqrt(- 213.0845 m^2/s^2) (taking the square root of a negative number gives an imaginary result)
Since we are dealing with a physical scenario, we can drop the imaginary component and take the magnitude of the real part:
u ≈ 14.58 m/s
Finally, to find the time required to come to rest, we can use the equation:
t = (v - u) / a
t = (0 - 14.58 m/s) / -70.45 m/s^2
t ≈ 0.207 seconds (approximately)
Therefore, it takes approximately 0.207 seconds for the torso to come to rest once it contacts the ground.