# Force at the end of a Rod at rest on an inclined surface

• so_gr_lo
In summary: There is no need to find the lever arm. That information is already in the diagram.No, you don't need to find the lever arm. That information is already in the diagram.
so_gr_lo
Homework Statement
End B of uniform rod of length 1.5 weight 30N rests against smooth surface inclined at 70°. Other end, A, of rod rests on rough ground at 50°. Find the magnitude of the force acting at B.
Relevant Equations
moment = F x d
Question diagram, attempt at solution below

I need to cancel some of the terms in the moment equation but a not sure which ones to start with. I don’t know μ so can not calculate FA, so should probably substitute FA = RB2.

I find your diagram and equations a bit confusing. Realize that at B, the force exerted by the surface on the rod must be perpendicular to the surface (because the surface is smooth).

so_gr_lo said:
Homework Statement:: End B of uniform rod of length 1.5 weight 30N rests against smooth surface inclined at 70°. Other end, A, of rod rests on rough ground at 50°. Find the magnitude of the force acting at B.
Relevant Equations:: moment = F x d

Question diagram, attempt at solution below
View attachment 274416
View attachment 274417
moment = F x d is only true if F and d are perpendicular, i.e. if d is the length of the 'lever-arm' as shown here: https://slideplayer.com/slide/14635...Scalar+Analysis+Mo+=+F+d+Mo+=+F+(r+sin+θ).jpg

One arrow from the rod's middle is the weight (W). No idea what the other arrow is.

You have marked the length incorrectly. The arrow heads should be level with the ends of the rod.

No idea what ##R_{B1}## is. But ##R_{B2}## is perpendicular to the (tilted) wall and there is no friction at B. So ##R_{B2}## is the required force.

There is no need consider horizontal/vertical forces. Take moments about point A. The clockwise moment produced by W equals the anticlockwise moment produced by ##R_{B2}##. That gives you a single equation from which to work out ##R_{B2}##.

You just need to use some simple trig’ to work out the lever -arm distances (d) for W and for ##R_{B2}##.

Delta2 and Chestermiller
Steve4Physics said:
moment = F x d is only true if F and d are perpendicular, i.e. if d is the length of the 'lever-arm' as shown here: https://slideplayer.com/slide/14635159/90/images/9/Calculating+moment+Scalar+Analysis+Mo+=+F+d+Mo+=+F+(r+sin+θ).jpg

One arrow from the rod's middle is the weight (W). No idea what the other arrow is.

You have marked the length incorrectly. The arrow heads should be level with the ends of the rod.

No idea what ##R_{B1}## is. But ##R_{B2}## is perpendicular to the (tilted) wall and there is no friction at B. So ##R_{B2}## is the required force.

There is no need consider horizontal/vertical forces. Take moments about point A. The clockwise moment produced by W equals the anticlockwise moment produced by ##R_{B2}##. That gives you a single equation from which to work out ##R_{B2}##.

You just need to use some simple trig’ to work out the lever -arm distances (d) for W and for ##R_{B2}##.
Apart from the way RB2 is drawn (seems it it supposed to be horizontal), everything looks correct so far. Also, it would be clearer if the surface were drawn to extend beyond the rod.
But there are only three equations and four unknowns. As @Doc Al posted, the unused fact is that the angled surface is smooth.

@so_gr_lo, a useful trick with such problems is to look for forces that you neither know nor need to know. By carefully choosing the axis to take moments about you can often eliminate these and get straight to the key equation.

My initial attempt at the solution was based on a tutorial for a similar question, however this question had a rough wall, so I included unnecessary forces.

if I take moments about A I can ignore RA and F. Below is my second attempt, but I am not sure if the line I have drawn is the correct perpendicular distance from R to A.

so_gr_lo said:
My initial attempt at the solution was based on a tutorial for a similar question, however this question had a rough wall, so I included unnecessary forces.

if I take moments about A I can ignore RA and F. Below is my second attempt, but I am not sure if the line I have drawn is the correct perpendicular distance from R to A.

View attachment 274465
You know R is perpendicular to the surface, and you want to draw a line perpendicular to R, so...

So the length of the surface is the perpendicular distance between R and A?

so_gr_lo said:
So the length of the surface is the perpendicular distance between R and A?
No, I wanted you to think about the direction of that perpendicular distance in relation to the plane of the surface.

The perpendicular distance is parallel to the surface

so_gr_lo said:
The perpendicular distance is parallel to the surface
Right. So if that distance meets the line of action of R at C, what is angle CAB?

Steve4Physics
20°

Steve4Physics
so_gr_lo said:
20°
Can I add that if the bottom of the wall is P, then ∠PBC = 90° and the lever arm (AC) is parallel to PB. Also ∠PBA = 20°. An accurate diagram can often make it much easier to see what's going on.

So for R I get

Lnewqban
Finally. Thanks for the help.

## What is the force at the end of a rod at rest on an inclined surface?

The force at the end of a rod at rest on an inclined surface is the force exerted on the rod by the surface due to gravity. This force is also known as the weight of the rod and is equal to the product of the mass of the rod and the acceleration due to gravity.

## How is the force at the end of a rod affected by the angle of inclination?

The force at the end of a rod is directly proportional to the angle of inclination of the surface. As the angle of inclination increases, the force at the end of the rod also increases. This is because the component of the force due to gravity acting perpendicular to the surface increases as the angle increases.

## What is the relationship between the force at the end of a rod and the length of the rod?

The force at the end of a rod is inversely proportional to the length of the rod. This means that as the length of the rod increases, the force at the end decreases. This is because the weight of the rod is distributed over a longer distance, resulting in a smaller force at the end.

## How does the force at the end of a rod change when the surface is frictionless?

If the surface is frictionless, the force at the end of the rod will be equal to the weight of the rod. This is because there is no opposing force from friction, so the entire weight of the rod is exerted on the end of the rod.

## Can the force at the end of a rod be greater than the weight of the rod?

No, the force at the end of a rod cannot be greater than the weight of the rod. The weight of the rod is the maximum force that can be exerted on the end of the rod, and it can only be equal to or less than the weight of the rod.

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