87 F, 1.5 m A Pivot point 0.1 m FIGURE 1.5 A force applied to an erect person. 直 0 78°F A AR AR pri scn FB home end FB dn 6d 5. 0.1 m Pivot point FIGURE 1.5 A force applied to an erect person. uscles The opposite restoring torque Tw due to the person's weight is T, =W×0.1m (1.2) W Assuming that the mass m of the person is 70kg, his weight W is ing W =mg=70 × 9.8 = 686 newton (N) (1.3) g Tip-Toe (Here g is the gravitational acceleration, which has the magnitude 9.8m/sec².) The restoring torque produced by the weight is therefore 68.6 newton-meter (N-m). The person is on the verge of toppling when the magnitudes of these two torques are just equal; that is, T, =T, or nic Posture W iction ing at an Fa × 1.5m=68.6N – m (1.4) on at the Hip Therefore, the force required to topple an erect person is 68.6 Fa= 1.5 e Fin of a 45.7N (10.31b) (1.5) cribology . Oral ribology Actually, a person can withstand a much greater sideways force without losing balance by bending the torso in the direction opposite to the applied 2. Hair cribology 3. Biotribology Textiles cises er 3: Translational 78°F ENG W re to search TO
87 F, 1.5 m A Pivot point 0.1 m FIGURE 1.5 A force applied to an erect person. 直 0 78°F A AR AR pri scn FB home end FB dn 6d 5. 0.1 m Pivot point FIGURE 1.5 A force applied to an erect person. uscles The opposite restoring torque Tw due to the person's weight is T, =W×0.1m (1.2) W Assuming that the mass m of the person is 70kg, his weight W is ing W =mg=70 × 9.8 = 686 newton (N) (1.3) g Tip-Toe (Here g is the gravitational acceleration, which has the magnitude 9.8m/sec².) The restoring torque produced by the weight is therefore 68.6 newton-meter (N-m). The person is on the verge of toppling when the magnitudes of these two torques are just equal; that is, T, =T, or nic Posture W iction ing at an Fa × 1.5m=68.6N – m (1.4) on at the Hip Therefore, the force required to topple an erect person is 68.6 Fa= 1.5 e Fin of a 45.7N (10.31b) (1.5) cribology . Oral ribology Actually, a person can withstand a much greater sideways force without losing balance by bending the torso in the direction opposite to the applied 2. Hair cribology 3. Biotribology Textiles cises er 3: Translational 78°F ENG W re to search TO
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Referring to Fig.1.5, compute the coefficient of friction at which the tendency of the body to slide and the tendency to topple due to the applied force are equal.
Please show me step by step and explain it.
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Transcribed Image Text:87
F,
1.5 m
A
Pivot point
0.1 m
FIGURE 1.5
A force applied to an erect person.
直 0
78°F A
AR
AR
pri scn
FB
home
end
FB
dn 6d
5.
Transcribed Image Text:0.1 m
Pivot point
FIGURE 1.5 A force applied to an erect person.
uscles
The opposite restoring torque Tw due to the person's weight is
T, =W×0.1m
(1.2)
W
Assuming that the mass m of the person is 70kg, his weight W is
ing
W =mg=70 × 9.8 = 686 newton (N)
(1.3)
g Tip-Toe
(Here g is the gravitational acceleration, which has the magnitude 9.8m/sec².)
The restoring torque produced by the weight is therefore 68.6 newton-meter
(N-m). The person is on the verge of toppling when the magnitudes of these
two torques are just equal; that is, T, =T, or
nic
Posture
W
iction
ing at an
Fa × 1.5m=68.6N – m
(1.4)
on at the Hip
Therefore, the force required to topple an erect person is
68.6
Fa=
1.5
e Fin of a
45.7N (10.31b)
(1.5)
cribology
. Oral
ribology
Actually, a person can withstand a much greater sideways force without
losing balance by bending the torso in the direction opposite to the applied
2. Hair
cribology
3. Biotribology
Textiles
cises
er 3: Translational
78°F
ENG
W
re to search
TO
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