Question

A 50 astronaut on Mars pushes a large crate at constant velocity as hard as she can without her feet slipping. The coefficient of static friction between her feet and floor is 0.8. The coefficient of kinetic friction between the crate and floor is 0.4.

(a) What is the largest crate mass she can push? Explain.

(b) If she applies the maximum push force she can exert, but the crate mass is reduced to half the value found in part, what will be the acceleration of the crate?

(c) Discussion question: if the force the astronaut exerts on the crate is (according to Newton’s 3^rd law) balanced by and equal and opposite force of the crate on the astronaut, why isn’t the net force zero making it impossible to push the crate?

Answer 1

part a:

let largest mass she can push is m kg.

as she is pushing with constant velocity, then the friction force and the pushing force are equal and in opposite direction.

maximum friction force on the crate=coefficient of kinetic friction*normal force

=0.4*m*g

where g=acceleration due to gravity on moon=1.6 m/s^2

for maximum value of m , it should match with maximum friction exerted on her from ground

which is equal to 50*1.6*0.8=64 N

then m*1.6*0.4=64

==>m=100 kg

so she can push a crate of maximum 100 kg mass.

part b:

if crate mass is reduced, friction force is reduced .

new friction force=0.5*100*1.6*0.4=32 N

then total force on the crate=64-32=32 N

acceleration of the crate=total force/mass=32/50=0.64 m/s^2

part c:

as the action and reaction forces are not acting on the same object, total force can not be added.

hence total force does not become zero.