You are Here: Home

# Mechanics/Dynamics Question

Announcements Posted on
Take our survey to be in with the chance of winning a £50 Amazon voucher or one of 5 x £10 Amazon vouchers 28-05-2016
First of all l don't understand the last line and don't know how to do part a of the question.

A horizontal force of magnitude 20N acts on a block of mass 1.5kg, which is in equilibrium resting on a rough plane inclined at 30 degrees to the horizontal. The line of action of the force is in the same vertical plane as the line of greatest slope of the inclined plane.

a) find the magnitude and direction of the frictional force acting on the block.

2. What does your working out look like?
3. Well... i don't really know when to start.
It says that the block is in equilibrium so i can't use uR.
Hmm so don't know what to do, grrr
4. (Original post by MathsBoy94)
It says that the block is in equilibrium so i can't use uR.
5. Draw the block with all the forces on (weight, friction, normal, applied force) then resolve parallel and perpendicular to the plane
6. In fact, I think you only need to resolve parallel to plane
7. i have, but friction is unknown
8. Ohhh Don't Worry I Got The Right Answer.
9. By resolving the perpendicular forces
10. (Original post by MathsBoy94)
First of all l don't understand the last line and don't know how to do part a of the question.

A horizontal force of magnitude 20N acts on a block of mass 1.5kg, which is in equilibrium resting on a rough plane inclined at 30 degrees to the horizontal. The line of action of the force is in the same vertical plane as the line of greatest slope of the inclined plane.

a) find the magnitude and direction of the frictional force acting on the block.

The below diagram might help:

11. (Original post by MathsBoy94)
By resolving the perpendicular forces
I think you only need to resolve parallel
12. (Original post by raheem94)
The below diagram might help:

Thanks, I did the first one.
On the second one it wanted me to find the normal reaction between the plane and the block.

The books says -2.7N
and i got 22.7N

Am not sure about the book
13. It is 22.7 if the horizontal force acts to the right as shown but it is -2.7 if it acts to the left
14. (Original post by leapingfrog)
It is 22.7 if the horizontal force acts to the right as shown but it is -2.7 if it acts to the left
I know but here is the diagram the book gives me next to the question
which shows the horizontal force pushing the block to the right.
Attached Thumbnails

15. It is 22.7
16. (Original post by MathsBoy94)
I know but here is the diagram the book gives me next to the question
which shows the horizontal force pushing the block to the right.
Its a mistake in the book. I have seen it. The answer at the back is -2.7 but the answer in the solution bank is 22.7.

See the image from the solution bank:

17. (Original post by raheem94)
Its a mistake in the book. I have seen it. The answer at the back is -2.7 but the answer in the solution bank is 22.7.

See the image from the solution bank:

If you know this then then you know in the second question after that there is also a mistake on question 10 can you please check to verify where i got 145.7 instead of 102 as it says in the book.
18. (Original post by MathsBoy94)
If you know this then then you know in the second question after that there is also a mistake on question 10 can you please check to verify where i got 145.7 instead of 102 as it says in the book.
The answer for it is 102N.

Both book and solution bank say the answer to be 102 and i also did it my self to ensure it, i get 102.

Post your working, so that i can correct you.
19. (Original post by raheem94)
The answer for it is 102N.

Both book and solution bank say the answer to be 102 and i also did it my self to ensure it, i get 102.

Post your working, so that i can correct you.
I got T-22gsin35-(0.125x22gcos35) = 0
20. (Original post by MathsBoy94)
I got T-22gsin35-(0.125x22gcos35) = 0

Why did you took the friction to be negative?

The particle is at the point of sliding down the plane, so the friction should act upwards because friction always opposes motion.

Do you get it?

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 1, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Don't be a half-term hermit

How to revise this week and still have a life

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams