COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 2, Problem 103P
To determine

The magnitude and direction of force exerted on the tibia by the femur.

Expert Solution & Answer
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Answer to Problem 103P

The magnitude of force exerted on the tibia by the femur is 281N_, and the direction is 39.7°below the horizontal to the right_.

Explanation of Solution

The free body diagram of the tibia is shown in the Figure 1 below.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 2, Problem 103P

Let positive x direction be along the direction of the length and positive y direction is the direction be along the perpendicular to the length of the leg.

From the free body diagram, write the expression for net force in the x direction.

    Fnet,x=mLgcos60.0°+mAgcos60.0°FPcos20.0°        (I)

Here, mLg is the weight of the lower leg, mAg is the weight of the ankle, FP is the force exerted by the femur on the tibia.

From the free body diagram, write the expression for net force in the y direction.

    Fnet,y=FPsin20.0°mLgsin60.0°mAgsin60.0°        (II)

Then the magnitude of the sum of the all forces on the tibia can be calculated as follows.

    Fnet=Fnet,x2+Fnet,y2        (III)

The angle made by the Fnet with respect to the leg can be calculated as follows.

    θ=tan1(Fnet,yFnet,x)        (IV)

Conclusion:

Substitute, 5kg for mL, 3kg for mA, 9.80m/s2 for g, and 337N for FP in the equation (I).

    Fnet,x=(5kg)(9.80m/s2)cos60.0°+(3kg)(9.80m/s2)cos60.0°(337N)cos20.0°=277.467N

Substitute, 5kg for mL, 3kg for mA, 9.80m/s2 for g, and 337N for FP in the equation (II).

    Fnet,y=(337N)sin20.0°(5kg)(9.80m/s2)sin60.0°(3kg)(9.80m/s2)sin60.0°=47.36N

Substitute, 47.36N for Fnet,y, and 277.467N for Fnet,x in the equation (III).

    Fnet=(277.467N)2+(47.36N)2=281N

Substitute, 47.36N for Fnet,y, and 277.467N for Fnet,x in the equation (IV).

    θ=tan1(47.36N277.467N)=9.7°

The magnitude of the force exerted by the femur on the tibia have the magnitude as Fnet, since the leg is stationary and it is directed opposite to it. So the direction of the force exerted on the tibia by femur can be calculated as follows.

    30.0°+9.7°=39.7°below the horizontal to the right

Therefore, the magnitude of force exerted on the tibia by the femur is 281N_, and the direction is 39.7°below the horizontal to the right_.

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Chapter 2 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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