Question:What is thefriction force on the box? Given:box=125N Force by person on

Question

Question:What is thefriction force on the box?
Given:box=125N Force by person N at 30 deg angle The coefficient ofstatic frcion = 0.80
Found that thenormal force is 163N
Usingformula
Fs= fn Fs = 0.80 *163N Fs =130N
However this is notcorrect. What am I doing wrong? I thought this was theformula.
Please help I amstumped.
I will ratelifesaver if able to answer me fully.
Thanks.
This is a weird situationsince the answer should be 130N so that is why I contacted theinstructor and verified if this was correct.
The friction should be thehorizontal forces only so Fcos * N
which is 0.8cos(30) *163 =130

Explanation / Answer

Here it is. I finally found the picture for it. Summing up the forces in the vertical direction we have-75NSin(210)-125N+N1=0 or N1=.87.5N. (where f=N1). Thenf=(.80)(87.5)=70N

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