How Do You Calculate the Resultant Force on a Mass on an Inclined Plane?

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To calculate the resultant force on a 100g mass on a smooth inclined plane, the primary force acting is the component of gravitational force parallel to the incline, expressed as F = mg sin(θ). The incline's dimensions allow for the calculation of the angle θ, which is essential for determining the force. Since the board is smooth, friction is negligible, simplifying the analysis. The discussion emphasizes that the distance along the incline does not factor into the force calculation, as it pertains to work done rather than resultant force. Understanding these principles is crucial for exam preparation in physics.
nigelhowie
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The diagram shows a smooth wooden board 30cm long. One end is raised 15cm above the other. A 100g mass is placed on the board. The two forces acting on the 100 g mass are shown in the free-body force diagram.

what is the magnitude of the resultant force?
 

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What do you think? Hint: Find components of the weight parallel and perpendicular to the board.
 
ok erm..

i was thinking

mgxsinθ = Frictional force x x

because since the object is not moving, gravitational potential energy is equals to frictional force?

please help me my edexcel exams are drawing really near.

thanks
 
btw, x= 30cm
 
When it says 'smooth' it means that there's no friction.

The only force on it as far as i can gather is the mgsin(theta).

So work out the force on it down the plane and from there you can work out acceleration and such, you only ever have to worry about friction if it says 'rough' or gives you a coefficient.

Edit: Also, out of interest is it the M1 exam you're taking? If so i'd check out this website

http://math.mdsalih.com/Data/index.php?d=Edexcel+Mathematics/M1

Many past papers - great practise with mark schemes.

Good luck.
 
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nope I'm taking physics unit 1.

this topic is killing me.

thanks for the link anyway =)

ok. so, if there is no frictional force acting on the body,

so, basically, the only force acting on the object is mgxsin(theta)?

F = mgxsin(theta)

?
 
Correcto!

good luck :D
 
Hey, no it's not, sorry :P

no X there, the force isn't dependant on the distance travelled, sorry i didn't read it right :O

The force here would only be F = mg sin (theta).

I think the only reason why it gives you those measurements is so that you can work out the angle! Give it a go and see what you get.
 
Chewy0087 said:
Hey, no it's not, sorry :P

no X there, the force isn't dependant on the distance travelled, sorry i didn't read it right :O

The force here would only be F = mg sin (theta).

I think the only reason why it gives you those measurements is so that you can work out the angle! Give it a go and see what you get.


oh yea thanks!

since

work done = force x distance moved in direction of force

so, X must not be calculated because if X is in, we're calculating work done.

thanks! =)
 
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