A suitcase of weight mg = 450 is being pulled by a small strap across a level floor. The coefficient of kinetic friction between the suitcase and the floor =0.640
a) Find the optimal angle of the strap above the horizontal. (The optimal angle minimizes the force necessary to pull the suitcase at constant speed.)
My answer is 32.6...is that right? If not please tell me how you did it...
Oh my answer is right... can someone tell me how to calculate the minimum tension in the strap needed to pull the suitcase at constant speed? My answer for the tension is 534N and I know it's wrong...
Please
well, if you are pulling at an angle of 32.6 deg...
for constant velocity, the horizontal force =friction force
friction force=(mg-upwardforce)mu
= (450-tension*sin32.4)mu
that has to equal horizontal force, or
horizontalforce= tension*cos32.4
set them equal
tension*cos32.4=450*mu-tension*sin32.4
tension= 450mu/(cosT-sinT)
solve for tension.
How did you get the angle?
To find the optimal angle of the strap above the horizontal, we need to determine the angle at which the force necessary to pull the suitcase at a constant speed is minimized. The force required to pull the suitcase can be determined by analyzing the forces acting on it.
Let's break down the forces acting on the suitcase:
1. Weight (mg): The weight of the suitcase acts vertically downward and is given as 450 N.
2. Normal Force (N): The normal force acts perpendicular to the surface of the floor, and its magnitude is equal to the weight of the suitcase when it is at rest or moving at a constant speed. Therefore, the normal force is also 450 N.
3. Friction Force (Ff): The friction force acts in the opposite direction of motion and opposes the movement of the suitcase. The magnitude of the friction force can be calculated using the coefficient of kinetic friction (μk) and the normal force (N). The equation is Ff = μk * N.
To find the optimal angle, we need to consider the vector components of the forces acting on the suitcase. Let θ be the angle of the strap above the horizontal.
Horizontal Component of Weight = mg * sin(θ)
Vertical Component of Weight = mg * cos(θ)
The force required to pull the suitcase at a constant speed can be calculated as:
Force required = Horizontal Component of Weight + Friction Force
Since the force required to pull the suitcase is minimized at constant speed, the friction force must be equal to the horizontal component of the weight.
Therefore, we equate the two:
Ff = mg * sin(θ)
μk * N = mg * sin(θ)
μk * mg * cos(θ) = mg * sin(θ)
Canceling out mg on both sides, we get:
μk * cos(θ) = sin(θ)
Dividing both sides by cos(θ), we obtain:
μk = tan(θ)
Finally, we solve for θ by taking the inverse tangent (arctan) of μk:
θ = arctan(μk)
Substituting the given coefficient of kinetic friction (μk = 0.640) into the equation, we find:
θ = arctan(0.640)
Using a calculator, we can evaluate the inverse tangent to find the optimal angle of the strap above the horizontal.