Force Problem - How do i get the inclination?

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To determine the minimum force required to move a 10.0 kg object on a horizontal surface with a coefficient of friction of 0.7, one must consider the forces acting on the object, including friction and the applied force at an angle. The normal force is calculated to be 98N, and the friction force is 68.6N. The relationship between the applied force and the angle can be expressed using the equation Fcos(x) = 0.7(w - Fsin(x)), leading to the equation 10cos(x) + 7sin(x) for minimizing the force. Solving this gives an inclination angle of approximately 34.99 degrees for the applied force. Understanding these concepts through free body diagrams and force balance equations is essential for solving similar physics problems effectively.
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A 10.0 kg object is resting on a horizontal surface with coefficient of friction of 0.7. Determine the minimum force F that can move this object. At what inclination should force F be applied? Present your answer in nearest hundredths.

I don't know how where to start?, How do i get the inclination ? thanks for the help
 
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I'm pretty sure there's an equation for this and you already have half of the variables. A quick google search would show you the answer. Try searching for coefficient of friction and in this case it's probably static friction.
 
ok thanks!
 
rose123456789 said:
ok thanks!
Does that mean you found the answer?
Really you should be figuring this out for yourself using free body diagrams and applying the force balance equations. You'll learn more that way than looking the answer up. Besides, it often happens that there is some subtle difference between the problem you are trying to solve and the one you find lying around on the net.
 
oh no, I'm still figuring out the answer
 
haruspex said:
Does that mean you found the answer?
Really you should be figuring this out for yourself using free body diagrams and applying the force balance equations. You'll learn more that way than looking the answer up. Besides, it often happens that there is some subtle difference between the problem you are trying to solve and the one you find lying around on the net.

i'm still figuring out the answer
 
rose123456789 said:
oh no, I'm still figuring out the answer
Ok. Have you drawn yourself a force diagram? What forces do you have, and in what directions do they act?
 
haruspex said:
Ok. Have you drawn yourself a force diagram? What forces do you have, and in what directions do they act?
i calculated that the normal force is 98N and the friction force is 68.6N . and i don't know what is next
 
rose123456789 said:
i calculated that the normal force is 98N and the friction force is 68.6N . and i don't know what is next
just take angle as some 'x' and force as F and write the equations .u wil get as
Fcos x=0.7(w-F sinx)
solving this u wil get the eq. as 10 cos x+7 sin x should me max for the force to be minimum.
and solving x=34.99 degrees
 
  • #10
rohitjinnu16 said:
just take angle as some 'x' and force as F and write the equations .u wil get as
Fcos x=0.7(w-F sinx)
solving this u wil get the eq. as 10 cos x+7 sin x should me max for the force to be minimum.
and solving x=34.99 degrees

how did you get 10 cos x +7 sin x
 
  • #11
rohitjinnu16 said:
just take angle as some 'x' and force as F and write the equations .u wil get as
Fcos x=0.7(w-F sinx)
solving this u wil get the eq. as 10 cos x+7 sin x should me max for the force to be minimum.
and solving x=34.99 degrees
ohhh ok i got it now thanks
 
  • #12
rose123456789 said:
ohhh ok i got it now thanks
kk fn.
 
  • #13
rohitjinnu16 said:
kk fn.

hahahahaha thanks a lot
 
  • #14
rohitjinnu16 said:
kk fn.
ok I'm a bit lost could you explain how did you get 10 cos x +7 sin x
 
  • #15
rose123456789 said:
ok I'm a bit lost could you explain how did you get 10 cos x +7 sin x
F cos x =0.7(w-F sin x)
F cos x/0.7=w-F sin x
F((cos x/0.7)+sin x)=w
F(10 cos x+7 sin x)/7=w
F=(7w)/(10 cos x+7 sinx)
 
  • #16
rose123456789 said:
i calculated that the normal force is 98N and the friction force is 68.6N . and i don't know what is next
You have not allowed for the applied force being at some angle to the horizontal. What will that do to the normal force?
 
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