Please I need some help in Newtons Laws

[ISU20CpreE]

In the example i have A block (mass m1 = 22.1 kg) is on a surface inclined at = 33°. It is connected to a mass m2 = 11.2 kg by a massless cord passing over a pulley. The coefficients of friction are µk = 0.15 and µs = 0.25
It asks me for the magnitud of the friction.

After looking at this problem I solved for Fs= 45.4 & the Fk=27.24. So i said that Fk<Fs. Therefore there is no acceleration. I really dont know where to go from there to get the magnitud of this friction. That its not even moving the block.

Please any sugestions??

[ISU20CpreE]

In the example i have A block (mass m1 = 22.1 kg) is on a surface inclined at = 33°. It is connected to a mass m2 = 11.2 kg by a massless cord passing over a pulley. The coefficients of friction are µk = 0.15 and µs = 0.25
It asks me for the magnitud of the friction.

After looking at this problem I solved for Fs= 45.4 & the Fk=27.24. So i said that Fk<Fs. Therefore there is no acceleration. I really dont know where to go from there to get the magnitud of this friction. That its not even moving the block.

Please any sugestions??


What no answers!!!:cry:

[Andrew Mason]

In the example i have A block (mass m1 = 22.1 kg) is on a surface inclined at = 33°. It is connected to a mass m2 = 11.2 kg by a massless cord passing over a pulley. The coefficients of friction are µk = 0.15 and µs = 0.25
It asks me for the magnitud of the friction.

After looking at this problem I solved for Fs= 45.4 & the Fk=27.24. So i said that Fk<Fs. Therefore there is no acceleration. I really dont know where to go from there to get the magnitud of this friction. That its not even moving the block.
You have to determine the other forces on the block along (parallel to) the surface. If the net force greater than the maximum static friction force, the block will move. In that case, the friction is kinetic. Otherwise, friction is static. If it is static, you do not use the coefficient of friction to determine the static friction force. Static friction will be whatever is required to balance the forces so that the block does not move.

AM

[ISU20CpreE]

Ok so what i got is m_1gSin\theta-F_f-F_T=m_1a Thats the only forces I should have.

I have my F_{net}=85.46 and the F_f=29.49, So i think the block is moving. Then if its moving what I should do next??

[Andrew Mason]

Ok so what i got is m_1gSin\theta-F_f-F_T=m_1a Thats the only forces I should have.

I have my F_{net}=85.46 and the F_f=29.49, So i think the block is moving. Then if its moving what I should do next??I get a net sideways force of:

m_1gsin(33) - T = m_1gsin(33) - m_2g = 118 - 110 = 8 N.

Since this is less than the maximum force that static friction can provide, the block does not move. What is the magnitude of F_s?

AM

[ISU20CpreE]

I get a net sideways force of:

m_1gsin(33) - T = m_1gsin(33) - m_2g = 118 - 110 = 8 N.

Since this is less than the maximum force that static friction can provide, the block does not move. What is the magnitude of F_s?

AM

I think if I use F_s=\mu_s*F_n It will give me a magnitud would that be right.

[Andrew Mason]

I think if I use F_s=\mu_s*F_n It will give me a magnitud would that be right.No. You do not use the co-efficient of static friction to find the actual static friction force. You use it to find the maximum possible static friction force. The static friction force provides just enough force to balance the other forces so the block does not move (zero acceleration).

AM

[ISU20CpreE]

No. You do not use the co-efficient of static friction to find the actual static friction force. You use it to find the maximum possible static friction force. The static friction force provides just enough force to balance the other forces so the block does not move (zero acceleration).

AM

Im sorry about not understanding. I believe that the static friction if its maximum will be \mu_s*F_n but how would I know for the amount of \mu_s?

[Andrew Mason]

Im sorry about not understanding. I believe that the static friction if its maximum will be \mu_s*F_n but how would I know for the amount of \mu_s?The forces on the block sum to 0 since the block does not move.

m_1gcos\theta - m_2g + F_s = 0

Therefore:

F_s = - m_1gcos\theta + m_2g = 118 - 110 = 8 N.

AM