Friction question

@Coppa

For some reason your post doesn't display properly on my old machine so I had to look at the page source - quoted in case anyone else has issues.

"A block of mass 10 kg rests on a rough horizontal surface. The coefficient of friction between the mass and the surface is 0.6. Calculate the magnitude of a horizontal force which just moves the block.Calculate the magnitude of a force acting at 30\u00b0 above the horizontal which just moves the block.Give you answers to 3 s.f. I managed to get the answer to the first part by using f=\"mew\" * Reaction force \nwhich is 98*0.6= 58.8 ( which is right) \nFor the second part of the question I used Fcos30=58.8 therefore 58.8\/cos30 = F F = 67.9 but apparantly this is wrong... can anyone tell me what I'm doing wrong ?"

For the second part, the vertical reaction force is no longer 10g - it's been reduced by a component of the force F.

(edited 3 years ago)

Friction question

ghostwalker

17

@Coppa

For some reason your post doesn't display properly on my old machine so I had to look at the page source - quoted in case anyone else has issues.

"A block of mass 10 kg rests on a rough horizontal surface. The coefficient of friction between the mass and the surface is 0.6. Calculate the magnitude of a horizontal force which just moves the block.Calculate the magnitude of a force acting at 30\u00b0 above the horizontal which just moves the block.Give you answers to 3 s.f. I managed to get the answer to the first part by using f=\"mew\" * Reaction force \nwhich is 98*0.6= 58.8 ( which is right) \nFor the second part of the question I used Fcos30=58.8 therefore 58.8\/cos30 = F F = 67.9 but apparantly this is wrong... can anyone tell me what I'm doing wrong ?"

For the second part, the vertical reaction force is no longer 10g - it's been reduced by a component of the force F.

(edited 3 years ago)

Reply 1

Coppa

4

Original post by ghostwalker

@Coppa

For some reason your post doesn't display properly on my old machine so I had to look at the page source - quoted in case anyone else has issues.

"A block of mass 10 kg rests on a rough horizontal surface. The coefficient of friction between the mass and the surface is 0.6. Calculate the magnitude of a horizontal force which just moves the block.Calculate the magnitude of a force acting at 30\u00b0 above the horizontal which just moves the block.Give you answers to 3 s.f. I managed to get the answer to the first part by using f=\"mew\" * Reaction force \nwhich is 98*0.6= 58.8 ( which is right) \nFor the second part of the question I used Fcos30=58.8 therefore 58.8\/cos30 = F F = 67.9 but apparantly this is wrong... can anyone tell me what I'm doing wrong ?"

For the second part, the vertical reaction force is no longer 10g - it's been reduced by a component of the force F.

Thanks for the help, I'm still unsure as how to calculate the new resultant force, since F is dependent on the friction which is it self dependent on the reaction force .

Reply 2

ghostwalker

OP

17

Original post by Coppa

Thanks for the help, I'm still unsure as how to calculate the new resultant force, since F is dependent on the friction which is it self dependent on the reaction force .

Start by resolving vertically; this will give you an equation in F, the applied force, and R the reaction.

Similarly resolve horizontally to get another equation.

And solve simultaneously.

A diagram might be useful. Post working if you get stuck.

Reply 3

Coppa

4

Thank you I managed to do it now !

Pls send it

Original post by PushTheA*

Pls send it

This thread is a year old and the OP hasn't logged on since last October!

It's probably a good idea to start your own thread with an example of your working if you need some help to complete the question :smile:

Reply 6

Nathaniel911

12

I know this thread is very old, ancient almost, but I thought I'd help any people still wanting the answer:

So you know that this angle is 30 degrees to the horizontal so resolving upwards we get:
R(^): R + Tsin30 - 10g = 0 so we now know that R is going to be equal to 10g - Tsin30

Because of this fact, we can now sub in the value of R into the friction equation knowing the coefficient of friction to be 0.6.
0.6(10g - Tsin30) = friction

For the block to be in equilibrium we need to equate the friction to the forward force of Tcos30
Simplfying you should get something similar to: Tcos30 + 0.6Tsin30 = 0.6x10g
Then by factoring out T you can equate T = 58.8/(cos30+0.6sin30)

Through solving you get 50.42771779 and the question specifically asks for 3.s.f therefore I input the answer as 50.4N which was correct!