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Anarel [89]
3 years ago
5

You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi

cular to the ground while the forearm is parallel to the ground. The male’s forearm is 26.5 cm long and weighs 1.42 kg. The man’s hand weighs 0.54 kg and is 19 cm long. The center of masses for the forearm and hand are 43% and 50% of the extremity lengths from the proximal joint, respectively.
You may assume that only the bicep generates the force, the angle between the bicep and forearm is 75 degrees, and the bicep insertion is 5cm from the elbow joint.

a) Draw a complete free body diagram of the above described activity. (5 pts)

b) Calculate the elbow moment due to external load and body segments weight. (5 pts)

c) Calculate the elbow contact force and force exerted by the bicep. (5 pts)

Engineering
1 answer:
cestrela7 [59]3 years ago
5 0

Answer:

Explanation:

The detailed analysis, with free body diagram and step by step calculations with appropriate substitution is as shown in the attached files.

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Part B Now push the plate down and observe what happens on the shore. Repeat the action three to four more times, and watch the
wolverine [178]

Answer:

A tsunami's trough, the low point beneath the wave's crest, often reaches shore first. When it does, it produces a vacuum effect that sucks coastal water seaward and exposes harbor and sea floors. As the tsunami approaches water is drawn back from the beach to effectively help feed the wave. In a tide the wave is so long that this happens slowly, over a few hours.

Explanation:

4 0
3 years ago
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this e
liubo4ka [24]

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

​

Explanation:

thats all you said

7 0
2 years ago
Read 2 more answers
Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C
Kruka [31]

Answer:

(4.5125 * 10^-3 kg.m^2)ω_A^2

Explanation:

solution:

Moments of inertia:

I = mk^2

Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2

Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2

Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2

Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.

r_A=50 mm

r_1 =100 mm

r_2 =50 mm

r_C=150 mm

At the contact point between gears A and B,

r_1*ω_b = r_A*ω_A

ω_b = r_A/r_1*ω_A

       = 0.5ω_A

At the contact point between gear B and C.

At the contact point between gears A and B,

r_C*ω_C = r_2*ω_B

ω_C = r_2/r_C*ω_B

       = 0.1667ω_A

kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2

                           =(4.5125 * 10^-3 kg.m^2)ω_A^2

6 0
3 years ago
Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth
lubasha [3.4K]

Answer:50 , 20

Explanation:

Given

Diametrical Pitch\left ( P_D\right )=\frac{T}{D}

where T= no of teeths

D=diameter

module(m) of gears must be same

m=\frac{D}{T}=\frac{1}{P_D}=0.1

Let T_1 & T_2be the gears on two gears

Therefore Center distance is given by

m\frac{\left ( T_1+T_2\right )}{2}=3.5

thus

0.1\frac{\left ( T_1+T_2\right )}{2}=3.5

T_1+T_2=70----1

and Velocity ratio is given by

VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear}

2.5=\frac{T_1}{T_2}----2

From 1 & 2 we get

T_1=50, T_2=20

6 0
3 years ago
Which of these is a scientific question
Paraphin [41]
The answer is B: How long does it take for bread dough to ferment
7 0
2 years ago
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