HW 10 - Part B Begin Date: 3/15/2018 12:01:00 AM - Due Date: 3/24 2018 11:59:00
ID: 2030343 • Letter: H
Question
HW 10 - Part B Begin Date: 3/15/2018 12:01:00 AM - Due Date: 3/24 2018 11:59:00 PMEnd Date: 3/31/2018 11:59:00 PM (20%) Problem 6: Consider the woman doing push-ups in the figure. She has a mass of 57.6kg, and the distance from her feet to her center of mass is 0.87 m, while the distance from her feet to her hands is 1.5 m CG Otheexpertta.com @ 25% Part (a) what force in newtons should the woman in the figure exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. 25% Part (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.95 cm, and she exerts force on the floor at a horizontal distance of 19 cm from the elbow joint. Calculate the magnitude of the force in newtons for each triceps muscle Grade Summary Deductions 6 100 Submissions Attempts remaining:9 (00 per attempt) detailed view cotan asinacos0 atan acotansinhO coshtanhO cotanh0 006 END Degrees O Radians BACKSPAC Submit Hint I give up Hints: 296 deduction per hint Hints remaining2 Feedback: deduction per feedback. -& 2596 Part (c) How much work in joules does she do if her center of mass rises 0.31 m? ilà 25% Part (d) What is her useful power output in watts. if she does 25 pushups in one minute?Explanation / Answer
Given,
m = 57.6 kg ; a = 0.87 m ; b = 1.5 m ;
b)d = 1.95 cm ; D = 19 cm
sum of the torques acting must be zero.
-Fp x 0 - w x a + 2 x Fr x 1.5 = 0
Fr = 57.6 x 9.81 x 0.87/2 x 1.5 = 163.866 N
Hence, Fr = 163.866 N
c)h = 0.31 m
W = m g h = 57.6 x 9.81 x 0.31 = 175.18 J
Hence, W = 175.18 J
d)P = E/t
for 25 push ups, E' = nE = 25 x 175.18 = 4379.5
P = 4379.5/60 = 72.99 W
Hence, P = 72.99 W (=73 W)
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