The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
loor on each hand and each foot, assuming that the person holds this position.
1 answer:
Complete Question
The complete question is shown on the first uploaded image
Answer:
Force on each hand is 196.22 N
Force on each foot is 95.8 N
Explanation:
In order to get a better understanding of this question let us explain some concepts
Normal Force:
We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.
Torque:
We can define torque as the moment of forces that tends to produce or cause rotation
From the question we are given that
Weight of body is (W) = 584 N
The normal force on both hands (Ha) = ?
The normal force on both legs (Lg) = ?
Looking at the diagram the person is at equilibrium so
584 = Ha + Lg
an also this mean that torques acting on the body is balanced
So, 0.410 Ha = 0.840 Lg
Making Lg the subject of formula in the equation above we
Lg = 0.4881 Ha
Considering the first equation and replacing Lg with this recent equation we have
584 = Ha + 0.4881 Ha
Therefore Ha = 392.44 N
This value obtained is for both hands for each hand we divide by 2
Therefore we have for each hand =196.55 N
Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have
Lg = 0.4881 × 392.44
= 191.22 N
The value above is the force on both legs to obtain the force on each leg we have
191.22/2 = 95.8 N.
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Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:
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Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):
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hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L