A foot strikes the floor with a force of 520N at an angle of 20 degrees. what is the friction force exerted by the foot?

Friction force on the ground is 488.6 N

Explanation:

As we know that the force of reaction which is along the horizontal surface in tangential direction is known as Friction force while other perpendicular component of the force is Normal force

Here we know that force is inclined at 20 degree with the horizontal

so we have

here we know

F = 520 N

so we have

a straight line sloped

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm