How do forces on a hippo change on a steeper hill?

[yosup231]

I feel like i basically know how to do this question but I am not 100% sure

A 1250 kg sliperry hippo slides down a mud covered hill included at an angle at 18 degrees to the horizontal. a) if the coefficient of sliding friction between the hippo and the mud is .09, what force of friction impedes the hippo's motion down the hill? b) if the hill were steeper, how would this affect the coefficient of sliding friction?

for a i think I am suppose to do (sin 18)(1250kg x 9.8m/s^2)(.09)= 347.64N but I am not sure if I am suppose to use cosine

for b i think htecoefficient of sliding friction would remain the same but I am not 100%

can someone help?

 

[Astronuc]

Friction is proportional (by virtue of the coefficient of friction) to the 'normal' force applied to the surface.

The weight (mg) of an object points downward (and perpendicular to the horizontal).

One needs to determine the normal force of the hippo on the slope.

What happens to the normal force when the angle increases?

 

[yosup231]

am i right?

i may be wrong but as the angle increases does the normal force decrease there for the proper equation to use is
(cos 18)(1250kg x 9.8m/s^2)(.09)= 1048.54N

 

[Jelfish]

That's correct!