# Negative Values of the Frictional Force?

I don't know if you'd find it helpful, but I like to start from Newton's law for this type of problem.

Net Force = Mass x Acceleration.

You've found the acceleration. The mass is 100/g kg.

The net force is the applied force less the frictional force = 40 - 100.μ N (where μ is the frictional coefficient to be found).

I get the same negative answer you did originally, too.

kuruman
Homework Helper
Gold Member
I just think it shows a lack of understanding that it was there in the first place.
Perhaps. Although not as simple as zero, 100 N cos90o is, nevertheless, the horizontal component of the weight.

Perhaps. Although not as simple as zero, 100 N cos90o is, nevertheless, the horizontal component of the weight.
Usually as folks progress on in the study of physics, they learn that they can usually omit the horizontal component of weight. :)

kuruman
Perhaps. Although not as simple as zero, 100 N cos90o is, nevertheless, the horizontal component of the weight.
Actually, the more I think about it, the more I think I agree with you - that it might not show a lack of understanding. It could actually be an indication of a fairly thorough understanding of the fundamental concept.

kuruman