M1 - Chapter 4 Help

Original post by HAtestemur

Hi guys, can someone help me answer this question I got in a quiz yesterday?

5:
(b) A body of mass 7 kg lies on a rough plane which is inclined at 35⁰
to the horizontal. The angle of friction between the body and plane
is 25⁰. Find the least force that must be applied to the body, in a
direction parallel to and up the plane, in order to prevent motion
of the body down the plane.

The answer is 13.2N to 3 significant figures.

I have never come across an "angle of friction" before

What board do you do?

Reply 2

brianeverit

9

Original post by TenOfThem

I have never come across an "angle of friction" before

What board do you do?

The angle of friction is the angle between the normal to the plane and the combined friction force and normal reaction.

Reply 3

Original post by brianeverit

The angle of friction is the angle between the normal to the plane and the combined friction force and normal reaction.

Not a concept I am familiar with

Certainly not a concept used in AQA or Edexcel

Reply 4

The angle of friction is the angle to the horizontal of the slope on which a freestanding body would be on the point of slipping.

\tan\theta=\mu

(edited 10 years ago)

Reply 5

Original post by ghostwalker

The angle of friction is the angle to the horizontal of the slope on which a freestanding body will start to slip.

\tan\theta=\mu

Right - that makes more sense - especially as it gives me mu

Reply 6

Original post by TenOfThem

Right - that makes more sense - especially as it gives me mu

Note minor edit to previous.

Reply 7

Original post by ghostwalker

Note minor edit to previous.

yes - slight but essential difference :smile:

Reply 8

OP

Has anyone managed to figure out a way to solve this question guy? I still don't know how to approach this question.

Its from a book that I bought of amazon with loads of problems inside.

Reply 9

Original post by HAtestemur

Has anyone managed to figure out a way to solve this question guy? I still don't know how to approach this question.

Its from a book that I bought of amazon with loads of problems inside.

Start with a diagram and mark on the forces.

You should know what the coeff. of friction is by the past discussion.

Can you work out the fricional force?

Then just resolve parallel to the plane.

Reply 10

OP

I drew a diagram with a plane inclined at 35 degrees from the horizontal.
7g pointing straight down from the box, and R pointing perpendicular to the plane.
Friction is applied at 25 degrees from the horizontal.

This is the part I may be going wrong on, I also labeled a Force called X pointing horizontal and upward to the plane because we are trying to find the force applied in order to prevent it from moving downwards. Is this correct ?

and to resolve, I resolved R perpendicular to the plane:

R + Fsin25 - 7gcos35 = 0
So R= 7gcos35 - Fsin25

and resolving up the plane:

X - 7gsin35 - Fcos35 = 0

and I don't know where to go after this.

Reply 11

Original post by HAtestemur

I drew a diagram with a plane inclined at 35 degrees from the horizontal.
7g pointing straight down from the box, and R pointing perpendicular to the plane.
Friction is applied at 25 degrees from the horizontal.

Frictional force will be along the plane.

This is the part I may be going wrong on, I also labeled a Force called X pointing horizontal and upward to the plane because we are trying to find the force applied in order to prevent it from moving downwards. Is this correct ?

Not quite. The force, X, is parallel to the plane, not horizontal.

Note: If you have a technique for using "angle of friction" you may have a different method of doing this.

I'd just use tan "angle of friction" = coeff. of friction, and then use coeff. of friction in calculations.

(edited 10 years ago)

Reply 12

OP

Original post by ghostwalker

Frictional force will be along the plane.

Not quite. The force, X, is parallel to the plane, not horizontal.

Note: If you're used to using "angle of friction" you may have a different method of doing this.

I have never answered questions of this types ever. I do not get a thing about this.

Reply 13

Original post by HAtestemur

I have never answered questions of this types ever. I do not get a thing about this.

OK, then go with coeff of friction = tan....

What equations do you get now?

Reply 14

OP

Original post by ghostwalker

OK, then go with coeff of friction = tan....

What equations do you get now?

I erased everything and drew a new sketch, with Friction parallel and down the inclined plane, and for X parallel and up the plane.

so resolving perpendicular and upwards,

R - 7gcos35 = 0
so R = 7gcos 35

and parallel up the plane:

X - 7gsin35 - F = 0
so X = 7g sin 35 + F

F = μR
so F = (tan25)(7gcos35)
so F = 26.23N

from there I said that

X = 7gsin35 + 26.23
X = 65.62N

but the answer is 13.2N

so when finding X, if I did 7gsin35 - 26.23 Id be getting 13.2N as the answer.

but how is it minus and not addition?

Reply 15

Original post by HAtestemur

I erased everything and drew a new sketch, with Friction parallel and down the inclined plane, and for X parallel and up the plane.

so resolving perpendicular and upwards,

R - 7gcos35 = 0
so R = 7gcos 35

and parallel up the plane:

X - 7gsin35 - F = 0
so X = 7g sin 35 + F

F = μR
so F = (tan25)(7gcos35)
so F = 26.23N

from there I said that

X = 7gsin35 + 26.23
X = 65.62N

but the answer is 13.2N

so when finding X, if I did 7gsin35 - 26.23 Id be getting 13.2N as the answer.

but how is it minus and not addition?

Because friction should be acting up the slope

Friction acts against motion and the block would be trying to move down

The key here is that you are looking for the "least force"

Reply 16

OP