please help! physics doesn't click with me. thank you for your help!
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?
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To find the force applied to the handle, we need to consider the forces acting on the suitcase. In this problem, we have two vertical forces acting on the suitcase: the weight (mg) and the normal force (N). The weight is acting downward, and the normal force is acting upward.
First, let's find the weight of the suitcase. The weight is given by the formula weight = mass x gravitational acceleration (w = mg). The mass of the suitcase is given as 23 kg, and the gravitational acceleration is approximately 9.8 m/s^2. Therefore, the weight of the suitcase is:
w = mg = 23 kg x 9.8 m/s^2 = 225.4 N
Now, let's consider the vertical forces. The normal force (N) is equal in magnitude but opposite in direction to the weight, so the normal force is 155 N (upward).
Since the suitcase is being pulled with a constant speed, we know that the net force acting on it in the horizontal (x-axis) direction is zero. In other words, the applied force in the horizontal direction is equal in magnitude and opposite in direction to the frictional force.
Now, let's analyze the forces in the horizontal direction. The applied force (F) can be divided into two components: one acting in the horizontal direction and one acting in the vertical direction.
The vertical component of the applied force is F_y = F sinθ, where θ is the angle (25°) and F_y is the force acting in the upward direction.
Since the suitcase is being pulled with a constant speed, the vertical forces must be in equilibrium, which means the vertical component of the applied force must balance the weight and the normal force. Therefore, we can write the equation:
F_y = w + N
Substituting the known values:
F sinθ = 225.4 N + 155 N
F sinθ = 380.4 N
Now, we can solve for F:
F = 380.4 N / sinθ
F = 380.4 N / sin(25°)
F ≈ 863 N
Therefore, the force applied to the handle is approximately 863 N.