KIN335 Problem Set #2 Solutions (11/6/06)
#1. (Example person: 5 foot 10 inches tall and weighing 176 lb)
a. W =176 lb = 176 lb (4.4482 N/lb) = 782.8832 N =783 N
b. W = mg so m = W/g = 782.8832 N / (9.8 m/s2) = 79.88604 kg = 79.9 kg
c. W = mg = (79.88604 kg)⋅(1.6 m/s2) =127.8177 N =128 N
d. H = 5ft 10in. =(5⋅12+10) in. = 70 in⋅(2.54 cm)/(1 in)=177.8 cm =178 cm =1.78 m
e. p=mv =(79.88604kg)⋅(27.5 m/s) = 2196.8661 kg⋅m/s = 2200 kg⋅m/s
#2.
m = 73 kg, W=mg = (73 kg)⋅(9.8 m/s2) =715.4 N = body weight (BW)
Newton’s Second Law Method:
ΣF = ma, where a= ∆v/∆t = [0 – (–4.3 m/s)] / (0.008 s) = 537.5 m/s2
So ΣF = (73 kg)⋅(537.5 m/s2) = 39237.5 N=39200 N (avg. resultant force)
or ΣF = 39237.5 N⋅(1 BW / 715.4 N) = 54.84694 BW = 54.8 BW
ΣF = GRF – W, so GRF = ΣF + W
GRF = 39237.5 N + 715.4 N = 39952.9 N = 40000 N (avg. ground reaction force)
or GRF = 39952.9 N⋅(1 BW / 715.4 N) = 55.84694 BW = 55.8 BW
Impulse Momentum Relationship Method:
ΣF(∆t) = mvf – mvi , ΣF = (mvf – mvi)/(∆t)
ΣF = [(73 kg ⋅ 0 m/s) – (73 kg ⋅(–4.3 m/s))] / (0.008 s)
=(73 kg)⋅(4.3 m/s) / (0.008 s)= 39237.5 N=39200 N (avg. resultant force)
or ΣF = 39237.5 N⋅(1 BW / 715.4 N) = 54.84694 BW = 54.8 BW
FBD:
+
ΣF = GRF – W, so GRF = ΣF + W
GRF = 39237.5 N + 715.4 N = 39952.9 N = 40000 N (avg. ground reaction force)
or GRF = 39952.9 N⋅(1 BW / 715.4 N) = 55.84694 BW = 55.8 BW
W
GRF
From the FBD above the resultant force (ΣF) is equal to GRF–W. Therefore, GRF = ΣF + W. That
indicates the vertical ground reaction force is always one body weight more than the resultant force.
Think of it as the first body weight of the GRF is used up in counteracting gravity to make a zero
resultant force. Anything past one body weight in the GRF makes the resultant force positive.
Anything less than one body weight in the GRF makes the resultant force negative.
#3.
Use the exact same procedure as #2 above but increase ∆t from 0.008 s to 0.35 s.
Then, ΣF = 896.85714 N = 897 N (avg. resultant force)
or ΣF = 896.85714 N⋅(1 BW / 715.4 N) = 1.25364 BW = 1.25 BW
GRF = 896.85714 N + 715.4 N = 1612.25714 N = 1610 N (avg. ground reaction force)
or 1612.25714 N⋅(1 BW / 715.4 N) = 2.253644 BW = 2.25 BW
Change of the average resultant force : ∆(ΣF)= (ΣF)0.35s –(ΣF)0.008s = 896.85714 N – 39237.5 N
= –38340.64286 N= –38300 N or –53.6 BW
(continued next page)
Problem Set #2 Solutions
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(#3 continued):
Change of the vertical GRF : ∆GRF= GRF0.35s –GRF0.008s = 1612.25714 N – 39952.9 N
= –38340.64286 N = –38300 N or –53.6 BW
So, the result indicates that making a large increase in landing time reduces the average GRF a
large amount which reduces the possibility of injury.
#4. (See handwritten solutions attached.)
#5.
m = 52 kg so W = mg = (52 kg) ⋅ (9.8 m/s2) = 509.6 N
a. Scaling reading (575 N) is greater than her body weight (509.6 N).
b. The resultant force=ΣF = GRF – W = (Scaling reading – W) = (575 – 509.6) = 65.4 N
c. ΣF = ma so a =ΣF / m = 65.4 N / (52 kg) = 1.257692 m/s2 = 1.26 m/s2
d. The elevator is speeding up since the acceleration is positive and the CM movement direction
is positive (the elevator is moving upward).
#6.
m =160 lb ⋅ (4.4482 N/lb) = 711.712 N, 1 in = 2.54 cm = 0.0254 m, 1 Pa =1 N/m2
90% of body weight affects pressure on the contact area of 150 in2.
P = F/A = (0.9⋅160 lb) / (150 in2) = 0.96 lb/in2
or 0.96 lb/in2 (4.4482 N/lb)(1 in / 0.0254 m)2 = 6618.93484 N/m2 = 6620 Pa or 6.62 kPa
#7.
m = 110 kg, vi = 8.5 m/s, vf = 0 m/s, ∆t = 1.5 s
ΣF⋅(∆t) = mvf – mvi , ΣF = m⋅(vf – vi) / (∆t) = (110 kg)⋅(0–8.5 m/s)/(1.5 s) = –623.333 N = –623 N
or 623 N (in the opposite direction)
#8.
Conservation of linear momentum: m1u1 + m2u2 = m1v1 + m2v2
where m1= 90 kg, m2 = 69 kg, ui=u2=0 m/s, v2 = 5.8 m/s. Find v1.
90 kg (0 m/s) + 69 kg (0 m/s) = 90 kg (v1) + 69 kg (5.8 m/s)
v1 = – [69 kg / (90 kg)] ⋅ (5.8 m/s) = –4.4467 m/s = –4.45 m/s or 4.45 m/s (to the left)
#9. (See handwritten solutions attached)
Problem Set #2 Solutions
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Problem Set #2 Solutions
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Problem Set #2 Solutions
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