Is the Trigonometric Analysis of Conveyor Belt Forces Correct?

[Aariz]

Homework Statement

solve the problems

Relevant Equations

mgsinΘ-µmgcosΘ=F

 

[DaveC426913]

You must make an effort at the solution before we can help you.

 

[Aariz]

Isn't the total force meaning the force needed to get that on the top...but i am confused about the height ..is height needed .?

 

[erobz]

Isn't the total force meaning the force needed to get that on the top...but i am confused about the height ..is height needed .?

Start with a free body diagram of the package on the slope. Draw the box and draw vectors representing the forces acting on it.

 

[jbriggs444]

Isn't the total force meaning the force needed to get that on the top...but i am confused about the height ..is height needed .?

You had written down an equation for force:

mgsinΘ-µmgcosΘ=F

Apparently, this is the component of net force acting on the box in the direction parallel to the ramp, diagonally downward. This under the assumption that friction is at its limiting value (just on the verge of slipping).

The $mg \sin \theta$ term is for the component of gravity parallel to the ramp (left and down).

$mg \cos \theta$ would be the normal force from the ramp on the box -- equal and opposite to the component of gravity normal to the ramp. We multiply by $\mu$ to get the limiting value of friction parallel to the ramp (acting up and to the right on the box.).

I agree with this formula.

However, there is a problem. If the box is not slipping on the conveyor then it is moving at a steady velocity. What must the "total force" be to maintain a steady velocity?

 

[erobz]

I agree with this formula.

According to the angle labeled as $\theta$ in the diagram? I think the trig functions are in your description are flipped. $mg \sin \theta$ is the normal force, and the weight of the box parallel is to the slope is $mg \cos \theta$.

The $mg \sin \theta$ term is for the component of gravity parallel to the ramp (left and down).

$mg \cos \theta$ would be the normal force from the ramp on the box -- equal and opposite to the component of gravity normal to the ramp. We multiply by $\mu$ to get the limiting value of friction parallel to the ramp (acting up and to the right on the box.).

 

[jbriggs444]

According to the angle labeled as $\theta$ in the diagram? I think the trig functions are in your description are flipped. $mg \sin \theta$ is the normal force, and the weight of the box parallel is to the slope is $mg \cos \theta$.

Yes, flipped.