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?

In the first question, if the upward component of the pulling force is 155N

then Pulling force= 155/sin25

I do not understand the b)

155/sin(25) is not the correct answer

To find the force F applied to the handle of the suitcase, we can use the concept of equilibrium. Since the suitcase is being pulled with a constant speed, the net force acting on it must be zero.

Let's break down the forces acting on the suitcase:

1. Weight (mg): The weight of the suitcase is given by the mass (23 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). So, the weight is 23 kg * 9.8 m/s^2 = 225.4 N. The weight acts vertically downward.

2. Normal force (N): The normal force exerted on the suitcase is 155 N. The normal force acts perpendicular to the surface of contact (which is horizontal).

3. Force applied to the handle (F): This force is acting at an angle of 25° above the horizontal.

Since the suitcase is in equilibrium, the vertical components of the forces must balance each other, and the horizontal components of the forces must also balance each other.

Vertical components:
- Weight (mg) acts vertically downwards, so its vertical component is -mg * cos(theta), where theta is the angle above the horizontal.
- Normal force (N) acts perpendicular to the surface, so its vertical component is N * sin(theta).

Horizontal components:
- The force applied to the handle (F) has a horizontal component equal to F * cos(theta).

In equilibrium, the sum of the vertical components of the forces is zero, and the sum of the horizontal components of the forces is zero.

Sum of vertical forces:
-N * sin(theta) - mg * cos(theta) = 0

Sum of horizontal forces:
F * cos(theta) = 0

Solving these equations, we can find the force F applied to the handle.

For the given values:
- N = 155 N
- m = 23 kg
- theta = 25°
- g = 9.8 m/s^2

F * cos(25°) = 0
F = 0

Since the force F is zero, it means that no force is applied to the handle. The suitcase is being pulled with a constant speed, so the force needed to balance the weight is provided by the normal force.

Now, let's move on to the second question:

To find the time required for the torso to come to rest once it contacts the ground, we can use the concept of free fall motion and equations of motion.

When the torso contacts the ground, the force acting on it is the normal force, which is 261,500 N. This force opposes the downward motion of the torso and causes it to come to rest.

The equation to use is:

Final velocity (vf) = Initial velocity (vi) + Acceleration (a) * Time (t)

Since the torso is falling under gravity, the initial velocity is zero (start of free fall) and the final velocity is also zero (at rest).

So, the equation becomes:

0 = 0 + a * t

The acceleration (a) is given by the force divided by the mass of the torso:

a = F / m
= 261,500 N / 3720 kg

Now, we can substitute the values into the equation and solve for time (t):

0 = (261,500 N / 3720 kg) * t

0 = 70.2903 m/s^2 * t

Since the left side is zero, we know that the product on the right side must also be zero. Therefore, time (t) is equal to zero.

This means that the torso comes to rest instantaneously upon contacting the ground.