An adult person is holding their arm straight out to their side. Their kid is hanging from their wrist, dangling. The kid has a mass of 20 kg. The length of adult’s arm between the shoulder and the wrist from which the kid is hanging is 0.7 m. The mass of the arm is 2.5 kg, and its center of mass is halfway along its length. The deltoid muscle is supporting all of this, generating a force directed at 14° above the arm and toward the body’s centerline. The deltoid muscle attaches to the arm at 15 cm from the shoulder. Find the magnitude of the force generated by this muscle.

asked by Jessica

12 years ago

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1 answer

The sum of torques about the shoulder is zero =>

mg•0.35 +Mg•0.7 -F•sinα•0.15 =0
F= g(m•0.35 +M•0.7)/ sinα•0.15=
=9.8(2.5•0.35+20•0.7)/0.24•0.15=
=4049 N.

answered by Elena

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Question

An adult person is holding their arm straight out to their side. Their kid is hanging from their wrist, dangling. The kid has a mass of 20 kg. The length of adult’s arm between the shoulder and the wrist from which the kid is hanging is 0.7 m. The mass of the arm is 2.5 kg, and its center of mass is halfway along its length. The deltoid muscle is supporting all of this, generating a force directed at 14° above the arm and toward the body’s centerline. The deltoid muscle attaches to the arm at 15 cm from the shoulder. Find the magnitude of the force generated by this muscle.

1 answer

The sum of torques about the shoulder is zero =>

mg•0.35 +Mg•0.7 -F•sinα•0.15 =0
F= g(m•0.35 +M•0.7)/ sinα•0.15=
=9.8(2.5•0.35+20•0.7)/0.24•0.15=
=4049 N.

Ask a New Question or answer this question.

Elena · Accepted Answer

The sum of torques about the shoulder is zero => mg•0.35 +Mg•0.7 -F•sinα•0.15 =0 F= g(m•0.35 +M•0.7)/ sinα•0.15= =9.8(2.5•0.35+20•0.7)/0.24•0.15= =4049 N.