Class Management i Help 9.5 & 9.6 Applications Begin Date: 3/7/2018 12:01:00 AM-
ID: 2269384 • Letter: C
Question
Class Management i Help 9.5 & 9.6 Applications Begin Date: 3/7/2018 12:01:00 AM- -Due Date: 3/18/2018 11:59:00 PM End Date: 3/31/2018 11:59:00 PM (17%) Problem 5: Consider the woman doing push-ups in the figure. She has a mass of 40.7 kg, and the distance from her feet to her center of mass is 0.83 m, while the distance from her feet to her hands is 1.75 m. CG reaction Otheexpertta.com Grade Summary Potential Late work % Late Potential 50% 50% coso sino cotan() atan cosh0 acoso acotan) sinh0 asino Attempts remaining: (20% per attempt) detailed view tanh() O Degrees O Radians 1-cotanh( ) 0 - Hints:-% deduction per hint. Hints remaining: Feedback: deduction per feedback. 25% Part (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.9 cm, and she exerts force on the floor at a horizontal distance of 18.5 cm from the elbow joint. Calculate the magnitude of the force in newtons for each triceps muscle. 25% Part (c) How much work in joules does she do if her center of mass rises 0,34 m? 25% Part (d) what is her useful power output, in watts, if she does 25 pushups in one minute?Explanation / Answer
Solution :-
mass m = 40.7 kg
a = 0.83 m
b = 1.75 m
(a. ) from the figure
using force balance
w = Fp + Fr
40.7*9.81 = Fp + Fr
from torque balance about the foot
w*a = Fr*b
40.7*9.81*0.83= Fr*1.75
Fr = 189.367 N
hence force by each hand = Fr/2 = 94.683N
(b.) lever arm, l = 1.9cm
d = 18.5 cm
hence force per tricep muscle = F
F*1.9= 94.683*18.5
F = 921.9134 N
(c. ) d = 0.15 m
work done = mgh = 40.7*9.81*0.34 = 135.7508 J
(d.) for 25 pushups pre minute
power = 135.7508*25/60 = 56.5628 W