A 22 kg suitcase is being pulled with constant speed by a handle that is at an angle of 23 ∘ above the horizontal.
If the normal force exerted on the suitcase is 150 N , what is the force F applied to the handle?
I don't understand how the equations are suppose to be manipulated. Please help
I assume you mean by "the normal force" the force the floor is exerting.
Normal force=weight-forcepulling up
= mass*9.8N/kg - F*sin23=150N
F=(mass*9.8-150)/sin23 deg
To solve this problem, we need to break down the forces acting on the suitcase and analyze them using trigonometry.
First, let's consider the weight of the suitcase. The weight is given by the equation W = mg, where m is the mass of the suitcase (22 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the weight W = 22 kg × 9.8 m/s^2 = 215.6 N.
Next, let's consider the normal force. The normal force is the force exerted by a surface that is perpendicular to it. In this case, the normal force is acting in the vertical direction to counterbalance the weight of the suitcase. Therefore, the normal force is equal to the weight, which is 215.6 N.
Now, let's analyze the forces acting in the horizontal direction. We have the force F applied to the handle and the force of kinetic friction Fk. Since the suitcase is being pulled with constant speed, the force of kinetic friction is equal in magnitude and opposite in direction to the applied force. Therefore, Fk = F.
Finally, let's use trigonometry to find the horizontal component of the force applied to the handle. The horizontal component can be determined using the equation Fx = F × cos(theta), where theta is the angle between the force and the horizontal direction. In this case, theta is 23 degrees. So, the horizontal component Fx = F × cos(23∘).
Now, equating the forces in the horizontal direction, we have:
Fx = F × cos(23∘) = Fk.
Substituting the known values, F × cos(23∘) = 215.6 N.
To solve for F, divide both sides of the equation by cos(23∘):
F = 215.6 N / cos(23∘).
Using a calculator, we can evaluate cos(23∘) ≈ 0.921.
F ≈ 215.6 N / 0.921 ≈ 234.22 N.
Therefore, the force F applied to the handle is approximately 234.22 N.