A man with a mass of 120 kg sits on the lion's back. How much normal force does the ground exerts upwards on the lion? How much normal force does the lion exerts upwards on the man? What is the net force acting on the lion?

1 Answer
Meave60
Oct 1, 2015

The normal force exerted by the ground on the lion and the man is #"3200 N"#.
The normal force exerted by the lion on the man is #"1200 N"#.
The net force acting on the lion is zero.

Explanation:

The downward force of weight will be represented by #"W"#, and the normal force will be represented by #"F"_"N"#.

Normal force exerted by the ground on the man and lion.

Known/Given:
#m_"man"=120"kg"#
#m_"lion"=210"kg"#
#m_"total"=330"kg"#
#g=-9.81"m/s"^2"#

Equation
#"W"=mg"#
#"F"_"N"=-"W"#

Solution:
#"F"_"N"=-"W"=-(330"kg")(-9.81"m/s"^2)=3237.3"N"=3200"N"# with two significant figures.

Normal force exerted by the lion on the man.
The normal force is equal but opposite to the weight of the man.

#"F"_"N"=-"W"=-mg=-(120"kg")(-9.81"m/s"^2)=1177.2"N"=1200"N"# with two significant figures.

The net force acting on the lion is zero because the lion is not moving, therefore its acceleration is zero.

#F_"net"=W_"total"+"F"_"N"#, where #"W"# is the combined weight of the lion and man, and the normal force is the upward force the ground exerts on both the lion and the man.

#W_"total"=(330"Kg")(-9.81"m/s"^2)=-3200"N"#

#F_"N"=-W=-(-3200"N")=3200"N"#

#F_"net"=F_"N"+W=3200"N"+(-3200"N")=3200"N"-3200"N"=0"N"#

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