- #1

Like Tony Stark

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- 6

- Homework Statement
- The coefficient of static and dynamic friction are 0.45 and 0.3 respectively. Find the maximun value that ##m_2## could have so that the system is in equilibrium. Then, if ##m_2## is the double of the mass calculated previously, determine what's the acceleration of the system.

- Relevant Equations
- Newton's equations

I considered the downwards direction and left direction as negative. For ##m_1##, Newton's equations are:

##x) Fr + W_x - T=0##

##y) N - W_y =0##

For ##m_2##:

##y) T - W =0##

Then, if I replace the data, I get ##T=22.2 N## and then ##m_2=2.2 kg##.

With that, for the second question ##m_2=4.4 kg##, and then solving Newton's equations I get for ##m_1##

##x) 151.96 - T=20.a_x##

And for ##m_2## I get

##y) T -44=4,4a_y##.

Then ##a_x=a_y## and if I solve the system I get ##T=63.92 N## and ##a=4,39 m/s^2##

But I have some doubts with the sign of the acceleration and tension. The acceleration must be negative, because ##m_2## will be moving downwards and ##m_1## will be moving to the left.

##x) Fr + W_x - T=0##

##y) N - W_y =0##

For ##m_2##:

##y) T - W =0##

Then, if I replace the data, I get ##T=22.2 N## and then ##m_2=2.2 kg##.

With that, for the second question ##m_2=4.4 kg##, and then solving Newton's equations I get for ##m_1##

##x) 151.96 - T=20.a_x##

And for ##m_2## I get

##y) T -44=4,4a_y##.

Then ##a_x=a_y## and if I solve the system I get ##T=63.92 N## and ##a=4,39 m/s^2##

But I have some doubts with the sign of the acceleration and tension. The acceleration must be negative, because ##m_2## will be moving downwards and ##m_1## will be moving to the left.