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Homework Help: Need help on these problems soon

  1. Sep 13, 2009 #1
    Need help on these problems soon plz

    Sorry again for not follow the format

    1. The Occupational Safety and Health Administration (OSHA) suggests a minimum coefficient of static friction of μs = 0.50 for floors. If Ethan, who has mass of 55 kg, stands passively, how much horizontal force can be applied on him before he will slip on a floor with OSHA's minimum coefficient of static friction?

    55x9.80=588.6x.50=269.5??? I need to follow significant digits on all the problems

    2. A 7.9 kg chair is pushed across a frictionless floor with a force of 42 N that is applied at an angle of 22° downward from the horizontal. What is the magnitude of the acceleration of the chair?

    can you plz check my work here
    horizontal line=sin(22)(42/sin(68)=16.969
    =3.17 m/s^2
    Is this right? help greatly appriaciated

    3. A shipping container is hauled up a roller ramp that is effectively frictionless at a constant speed of 2.10 m/s by a 2250 N force that is parallel to the ramp. If the ramp is at a 24.8° incline, what is the container's mass?
    m=109.329 kg

    Are these correct, also check my significant digits Thanks so much
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 14, 2009 #2
    Re: Need help on these problems soon plz

    Hi Ion1776

    Question 1
    I think your answer is right, but you have to state it in 2 significant digits because the digits in the question have maximum 2 significant digits.

    Question 2
    If I understand your question correctly, the chair moves horizontally so you only consider the horizontal force. Find the horizontal force that acts on the chair then find the acceleration

    Question 3
    The acceleration is not 2.10m/s. It's the speed. The container moves with constant speed 2.10m/s, so can you find the acceleration?
    A 2250N force is needed to move the container up the ramp because there is another force that resist the motion, which acts downward and parallel to the ramp. You have to consider this force when applying [tex]\Sigma F=ma[/tex]

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