A woman at an airport is towing her 17.0 kg suitcase at constant speed by pulling on the strap at an angle θ above the horizontal. She pulls on the strap with a 26.0 N force, and the friction force on the suitcase is 14.0 N.
To find the angle θ at which the woman is pulling the suitcase, we can use the concept of vector components. First, let's break down the forces acting on the suitcase along the horizontal and vertical directions.
1. Along the horizontal direction:
The component of the pulling force in the horizontal direction is given by:
Fx = F * cos(θ)
where F is the pulling force (26.0 N) and θ is the angle above the horizontal.
2. Along the vertical direction:
The gravitational force acting on the suitcase is given by:
Fg = m * g
where m is the mass of the suitcase (17.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
The friction force acting on the suitcase opposes the motion, so it points in the opposite direction of Fx. Therefore, the vertical component of the pulling force is equal to the gravitational force:
Fy = Fg + Ff
where Ff is the friction force (14.0 N).
Fy = F * sin(θ)
Now that we have the equations for the horizontal and vertical components of the pulling force, we can solve for θ.
1. Rearrange the equation for Fx:
Fx = F * cos(θ)
F * cos(θ) = Fx
2. Rearrange the equation for Fy:
Fy = F * sin(θ)
F * sin(θ) = Fy
3. Divide the two equations to eliminate F:
(F * sin(θ)) / (F * cos(θ)) = Fy / Fx
sin(θ) / cos(θ) = Fy / Fx
tan(θ) = Fy / Fx
4. Take the inverse tangent (arctan or tan⁻¹) of both sides to solve for θ:
θ = tan⁻¹(Fy / Fx)
Plug in the values of Fy (Fg + Ff) and Fx to calculate θ.