Calculating Force: Bicycle & Block Mass, Friction Coefficients

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1.
A cyclist is attached to a rope which is looped over a pulley (disc of mass 10kg and radius 0.5m). The rope is then
attached to a hanging block of mass 5kg. What force must the cyclist generate such that the block will move with an
acceleration of 2.25 m/s2. Let’s make the assumption that only one wheel of the bicycle is necessary to generate the
force (we can ignore the front wheel) and that all the force will manifest itself in rolling friction. The mass of the
bicycle and the cyclist is 90kg. The surface has a coefficient of static friction of 0.65 and kinetic friction of 0.25.
mass of a wheel is 700g
diameter of a wheel is 622mm


2. I am having a difficult time drawing out the FBD for each piece. I know from summing the forces on each piece you can calculate the force the wheel needs to generate .3. I know for the block there is tension in the rope pointing up and mg pointing down. I know the disk is rotating clockwise and that it has some torque. I also know that there is friction on the bike along with the normal force and weight. The FBD of the disk is confusing me along with the problem.
 
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Can you draw the figure?
 
JoeSmith1013 said:
The FBD of the disk is confusing me along with the problem.
I assume the rope is horizontal from pulley to cyclist, and that the cyclist is on a horizontal road.
For the pulley, just treat the tensions as two forces applied at the perimeter, one vertically down to the mass and one horizontal to the cyclist. There will also be a force on the axle from the pulley's support, but you don't need to worry about that since you only care about the angular acceleration equation.
Post what equations you can.
 
I know alpha=T/I So I am guessing the force moving the box is the torque in the wheel of the bike.
 
I also know Icom of a disk is 1/2Mr^2
 
Confused on where to go from here
 
JoeSmith1013 said:
I know alpha=T/I So I am guessing the force moving the box is the torque in the wheel of the bike.
We're not discussing the bike yet. I was replying to your saying you were stuck on the FBD for the pulley. When drawing an FBD for a rigid body, you work only with the forces acting directly on that body. For the pulley, there are just three: the two tensions (which will be diffeent) and the force on the axle from the support.
Create unknowns for the forces as necessary. In terms of those, what torques act on the pulley?