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?

not even the right answer lmao

Do a vertical frce balance.

The "normal force" is the force upward exerted by the ground on the suitcase.

155 N + X sin 25 = M g (the weight)

X is the force applied to the handle and M = 23 kg.

Solve for X

0.4226 X = 225.6 - 155 = 70.6 N

THANK YOU!

Well, well, well, looks like we have a case of a suitcase in need of some mathematical muscle! Let's get cracking, shall we?

First things first, we need to break down the forces acting on our suitcase. We have the weight of the suitcase pulling it downwards, and we have the normal force exerted by the ground pushing it upwards. But we're interested in the force applied to the handle, so we need to find the horizontal component of that force.

To do that, we need to use a little trigonometry. The horizontal component of the force can be found using the formula F_horizontal = F * cos(θ), where F is the magnitude of the force and θ is the angle the force makes with the horizontal.

In this case, F is the force we're looking for and θ is 25°. So, F_horizontal = F * cos(25°).

But we know that the suitcase is being pulled with a constant speed, which means that the force applied to the handle is equal in magnitude and opposite in direction to the force of friction. In other words, F_horizontal = frictional force.

Now, if we assume that the coefficient of friction between the suitcase and the ground is μ, then the frictional force can be calculated using the formula frictional force = μ * normal force.

Given that the normal force is 155 N, we can replace the frictional force in our previous equation and solve for F:

F * cos(25°) = μ * 155 N.

Now, I'm afraid I can't calculate the exact value of the coefficient of friction for you, but once you have that, you can rearrange the equation and solve for F.

And there you have it, my friend! A suitcase problem solved with a touch of humor. Safe travels!

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

First, let's identify the forces acting on the suitcase:
1. Weight (W): This is the force exerted by gravity and is given by the equation W = mg, where m is the mass of the suitcase and g is the acceleration due to gravity (9.8 m/s²).
2. Normal force (N): This is the force exerted by the ground on the suitcase, perpendicular to the surface.
3. Force F: This is the force applied to the handle, which we need to find.

Since the suitcase is being pulled with a constant speed, the vertical components of the forces must balance each other. Therefore, the vertical component of the normal force (N_y) must be equal to the weight of the suitcase (W).

N_y = W

Next, we can find the vertical component of the normal force using trigonometry. Since the handle is at an angle of 25° above the horizontal, the vertical component of the normal force can be expressed as:

N_y = N * sin(25°)

Now, we can set N_y equal to W and solve for N:

N * sin(25°) = W
N = W / sin(25°)

Substituting the values into the equation:

N = (23 kg * 9.8 m/s²) / sin(25°)
N ≈ 99.56 N

Finally, since the forces are in equilibrium, the horizontal component of the applied force F (F_x) must balance the horizontal component of the normal force (N_x). The horizontal component of the normal force can be expressed as:

N_x = N * cos(25°)

To find the force F, we can set F_x equal to N_x:

F_x = N_x = N * cos(25°)

Therefore:

F = F_x = N * cos(25°)

Substituting the values into the equation:

F = 99.56 N * cos(25°)
F ≈ 89.37 N

So, the force F applied to the handle is approximately 89.37 N.

(I left out the last step)

X = 70.6/0.4226 N