Two masses are connected by a string on a pulley as shown in the figure where m1 = 4.05kg and m2 = 2.82kg. The coefficient of kinetic friction is 0.254. Assuming that we take the direction to the right as positive, what is the largest numerical value of the vector F to keep the tension of the string at zero?

Question

anonymous · Answer

Figure

anonymous · Answer

You have forces , sum them first

anonymous · Answer

m1g

anonymous · Answer

-m1g + .254m2g = -32.670456 N

anonymous · Answer

I would assume that Fx would be 32.67 N, but that's incorrect.

anonymous · Answer

Adding the forces as if they had the same sign, 46.71 N to the left.

anonymous · Answer

$$\sum F_x=f_x-0=m_1a_x $$
$$\sum F_y=ma_x=0-m_2g$$

anonymous · Answer

$$m_2a_y=-m_2g$$
$$a_y=-g$$

anonymous · Answer

since they're attached to each other they have the same acceleration

anonymous · Answer

soyou end up with $$m_1a_x=f_x$$

$$m_1(-g)=f_x$$

anonymous · Answer

I have tried that, but it is not the largest value for Fx.

anonymous · Answer

so you just plug in m1, and g and thats the largest force... it's theo nly force that will allowthe tension to be zero

anonymous · Answer

don't havea calculator

anonymous · Answer

Kinetic friction and the m2 block were not taken into consideration, therefore this answer is incorrect. You would be right if Fx was pulling only on m1, but in the figure, this is not the case.

anonymous · Answer

not sure what you're trying to say. m2 was taken into consideration . Fx is only pulling on the object m1 and not m2

anonymous · Answer

It's only pulling on m2, not m1.

anonymous · Answer

oh lol switch the m1 wit m2

anonymous · Answer

in my summation m1 is m2 in your diagram

anonymous · Answer

I see. Thanks for your time!