Force on sliding box.

  • Thread starter Tanya Back
  • Start date
  • #1
18
0
Hey

I am having trouble doing this question-->

A gardener exerts a force of 150 N [ 22 degrees below the horizontal] in pushing a large 18 Kg box a distance of 1.6 m. The coefficient of kinetic friction between the box and the floor is 0.55.
Question -->
Use Newton's law to determine magnitudes of the normal force and force of friction on the box?

Solution :

Fn= Fag perpendicular
= Fgcosa
= (9.8)(18)cos22
= 164 N

Ff= uFn
=0.55(164)
= 89.9 N
***********
My answers are wrong becasue the answer is suppose to be Fn = 230 N and Ff= 130 N
What did i do wrong? how can i fix it?
 

Answers and Replies

  • #2
298
0
Well first of all draw yourself a free-body diagram. The normal force points directly upwards. There are two forces that are pointing downwards. Since there is no up or down motion the normal force must be equal to the two.
 
  • #3
18
0
ohh k ...so Fg + 150 = Fn ?
 
  • #4
298
0
Nope, but you're getting closer. The force that the gardener applies is at an angle. So only the vertical component points downwards. Did you draw your free-body diagram?
 
  • #5
18
0
Oh.. Yah i drew the free-body diagram..and i got something like this -->

its like Fgcos22 + 150cos68 = Fn
 
  • #6
298
0
Why do you have the cos22 for the force of gravity? I think you might have drawn your free body diagram wrong. Can you show me a crude drawing of it looks like?
 
  • #7
18
0
OHHHHHH! opps ...i meant to write Fg + 150cos68= Fn
so it's like 174 + 56 = 230... Thank u sooo Much !!
 

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