For the above situation we will draw a force diagram first
Now as per force diagram we can see that there is a component of force parallel to the surface due to which sled is pulled forward.
Now this force must be counter balanced by friction force as the sled is moving with constant speed.
So if sled is moving with constant speed it will have net force ZERO
by Newton's II law
[tex]F_{net} = ma= 0[/tex]
so we can say
[tex]F_{friction} = Fcos\theta[/tex]
[tex]F_{friction} =80cos53[/tex]
[tex]F_{friction} = 48 lb[/tex]
So friction force on it is 48 lb.
