A sprinter performs drills where he brings one knee up to his chest and raises himself up on the ball of his foot on the supporting leg. He maintains this position, then slowly lowers the leg. When he is balanced on the ball of his foot, what is the force exerted by the Achilles tendon if the distance from the ball of his foot to his calcaneus (the back of his foot) is 20 cm, his weight is 70 kg and the force from his body weight acts 8 cm from the ball of his foot? Take g=10 m/s2 and the angle between the foot and group to be 30°.
1. Draw out a free body diagram. 2. Determine a Lever Arm for each force (the perpendicular distance from the axis to the line of action of the force). 3. Determine the component of the force that is perpendicular to the Lever Arm. In this problem this is not necessary. 4. The torque produced by a force is equal to the component of the force that is perpendicular to the lever arm, multiplied by the length of the lever arm. tendon=(x).20(cos30°)m bw=700N.08(cos8°)m=55.5Nm Torque is a vector so it is important to keep track of direction. If a force causes counterclockwise rotation, it is customary to define the torque as positive. Our force of the tendon acts counterclockwise (+). We state our force of body weight acts clockwise (-). 5. After finding all of the torques, it is important to write down whether the object is accelerating or in equilibrium based on the problem. In this problem we are in equilibrium, which means the sum of the torques equals 0 =(x).20(cos30°)m-55.5Nm= 0 (x).20(cos30°)m=55.5Nm (x)=320.2N
The diagonal AB makes two 45°-45°-90° triangles with the sides of a square. What is the area of the square if AB= 4root2 m?
Step 1: Draw the figure, making sure to indicate right angles. Step 2: Using a) the pythagorean theorem or b) the 45°-45°-90° special triangle ratio n : n : n root2 find the length of a side: If the hypotenuse is 4 root2 then the legs must be 4. So the length of the side of the square is 4. Step 4: Area of square = h*h = 4*4 = 16m(squared)
What are two reason HF is a weaker acid than HCl?
HF is a weaker acid than HCl because 1. The HF bond is shorter (and stronger) compared to the larger HCl. This means there is greater difficulty in donating that acid proton in HF than in HCl. 2. The charge of a fluoride anion is distributed over a much smaller radius than that of Cl(-), making it quite reactive. It is a less stable conjugate base (the base that is created when an acid donates a proton) than Cl(-). The free energy (ie, favorability) of the reaction is lower for F(-) than for Cl(-), and the "acid activity" of HCl is therefore more favorable, because Cl(-) anion is more stable. Re: conceptual challenges A common difficulty students have is acidity generally increases across a row in the periodic table, owing to the greater affinity of these atoms for a negative charge. This means the conjugate base is more stable moving across a row. An atom happier with a negative charge (more electronegative) is more likely to donate that proton and adopt that negative charge. Based on electronegativity alone, HF would appear to be more acidic. However, moving down a period, as atomic radii increase for atoms in the same group, the conjugate bases get more stable, to the end result that the most acidic behavior is seen not at the top right as you'd expect purely accounting for electronegativity, but bottom right.