SOLUTION: A 10 kg box is resting on a ramp inclined at 35° to the horizontal. Resolve the weight of the box into the rectangular vector components that keep the box at rest. Include a diagr

Algebra ->  Vectors -> SOLUTION: A 10 kg box is resting on a ramp inclined at 35° to the horizontal. Resolve the weight of the box into the rectangular vector components that keep the box at rest. Include a diagr      Log On


   

Question 1159990: A 10 kg box is resting on a ramp inclined at 35° to the horizontal. Resolve the weight of the box into the rectangular vector components that keep the box at rest. Include a diagram illustrating this situation.
Found 2 solutions by ikleyn, MowMow:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

The components of the weight are  


    W*sin(a) = 10*sin(35°) = 10*0.5733 kilograms = 5.733 kilograms = 5.733*9.81 newtons = 56.243 newtons PARALLEL the incline

and

    W*cos(a) = 10*cos(35°) = 10*0.8193 kilograms = 8.193 kilograms = 8.193*9.81 newtons = 80.376 newtons PERPENDICULAR to the incline.

Solved.

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Notice that kilograms are not used as the unit of weight (of force) since sixties
last century, been replaced by Newtons.

So your source is  EXTREMELY  outdated.

The numbers by @MowMow are incorrect (!)

Avoid use it (!)

@MowMow,  you need seriously update/refresh your knowledge (!) (!)

Had I calculate forces for engineering projects at my positions here in US,  as you do in your post,
I would be fired next day.

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Regarding a diagram,  open any textbook in Physics and find a diagram there.

Or look into the lesson
    - Using vectors to solve problems in Mechanics: Force,  Figures  3,  4  and  5
in this site.


Answer by MowMow(42) About Me  (Show Source):
You can put this solution on YOUR website!
Fv = 10Cos(55) and Fh = Fh/10 = Sin(35)

The box is kept at rest by a force of 5.74 N perpendicular to the ramp and by a friction of 8.19N parallel to the ramp