A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. If the normal force exerted on the suitcase is 150 N, what is the force F applied to the handle?

Up force from pulling= Fsin29

Normal force= mg-up force
150N=23*9.8-Fsin29
solve for F.

To find the force F applied to the handle, we can break down the forces acting on the suitcase into horizontal and vertical components.

1. Vertical Forces:
The weight of the suitcase acts vertically downwards. The weight can be calculated using the formula:
Weight = mass * gravitational acceleration (W = m * g)
Here, the mass of the suitcase is given as 23 kg, and the gravitational acceleration is approximately 9.8 m/s^2.
So, the weight of the suitcase is:
Weight = 23 kg * 9.8 m/s^2 = 225.4 N (approximately)

The normal force exerted on the suitcase is equal to the vertical component of the force F exerted on the handle. Since the suitcase is in equilibrium, the normal force is equal to the weight of the suitcase:
Normal force = Weight of the suitcase = 225.4 N

2. Horizontal Forces:
The horizontal component of the force F applied to the handle is responsible for maintaining the constant speed of the suitcase. This force can be calculated using trigonometry.
The horizontal component can be found using the equation:
F_horizontal = F * cos(theta)
Here, theta (θ) is the angle between the handle and the horizontal, which is given as 29°.
The cosine of 29° is approximately 0.866.
So, the horizontal component of the force F is:
F_horizontal = F * 0.866

Since the suitcase is moving at a constant speed, the net force acting on it in the horizontal direction is zero. Therefore, the horizontal component of the force F is equal to the force of friction.

3. Frictional Force:
The force of friction can be calculated using the formula:
Frictional force = μ * Normal force
Here, μ (mu) is the coefficient of friction between the suitcase and the surface it is being pulled on.

Since the suitcase is being pulled with a constant speed, the force of friction is equal to the horizontal component of the force F:
Frictional force = F_horizontal
μ * Normal force = F * 0.866

4. Combining Equations:
Equating the expressions for the frictional force gives:
F * 0.866 = μ * Normal force

Substituting the given values:
150 N * 0.866 = F
F = 129.9 N (approximately)

Therefore, the force applied to the handle is approximately 129.9 N.