BIO Peak Pedaling Torque The downward force produced by the quadriceps muscles during the power stroke of bicycle-pedaling motion is shown in Figure 11-67 as a function of the crank angle ϕ (see Figure 11-39 .The force from these muscles decreases linearly, but the torque depends on the crank angle according to τ = rF sin (180° − ϕ ). Use the information in the graph, together with a computer spreadsheet to find the angle ϕ at which the pedaling torque produced by the quadriceps muscle is a maximum. (Note that the actual torque applied to the crank is a result of the action of many muscles in addition to the quadriceps). Figure 11-67 Problem 89
BIO Peak Pedaling Torque The downward force produced by the quadriceps muscles during the power stroke of bicycle-pedaling motion is shown in Figure 11-67 as a function of the crank angle ϕ (see Figure 11-39 .The force from these muscles decreases linearly, but the torque depends on the crank angle according to τ = rF sin (180° − ϕ ). Use the information in the graph, together with a computer spreadsheet to find the angle ϕ at which the pedaling torque produced by the quadriceps muscle is a maximum. (Note that the actual torque applied to the crank is a result of the action of many muscles in addition to the quadriceps). Figure 11-67 Problem 89
BIO Peak Pedaling Torque The downward force produced by the quadriceps muscles during the power stroke of bicycle-pedaling motion is shown in Figure 11-67 as a function of the crank angle ϕ (see Figure 11-39.The force from these muscles decreases linearly, but the torque depends on the crank angle according to τ = rFsin (180° − ϕ). Use the information in the graph, together with a computer spreadsheet to find the angle ϕ at which the pedaling torque produced by the quadriceps muscle is a maximum. (Note that the actual torque applied to the crank is a result of the action of many muscles in addition to the quadriceps).
Q) A simplified model as shown in figure below of a planar robotic am. You may
assume that each segment has length 1.5 m, and the applied force is 250 N. 0, =
35" and 02 = 0; = 20°.
Calculate the moment that the force, F, exerts about the point A.
F
0:
A
Newton's second law for linear motion is mà The equivalent form for rotational motion is - là, where is torque, I stands for the moment of inertia, and is the angular acceleration. The effectiveness of the
applied torque is determined by:
O The amount of Force applied
The distance of the applied force from the axis of rotation.
O The angle between the applied force vector Fand the vector distance from the axis of rotation to the applied point of the force
O All of the above
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A force F = 29 N i acts on an object at a point
(xo, Yo) = (4.6 m, 4.7 m) as shown in the figure.
y
Next
Yo
F
F
X
What is the magnitude of the torque generated by this
force about the origin?
What is the magnitude of the torque generated by this
force about the point (x, y) = (2.1 m, 2.2 m)?
Suppose the object is free to rotate about the z axis. If
the object has a moment of inertia | = 33.0 kg*m² for
rotation about the z axis, what is the magnitude of the
angular acceleration of the object due to the application
of the force F?
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