Two blocks connected by a rope of negligible mass are being dragged by a force at a 22° angle above horizontal (see figure below). Suppose F = 89.0 N, m1 = 14.0 kg, m2 = 22.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.128

Question

anonymous · Answer

http://www.webassign.net/userimages/5-p-047.gif?db=v4net&id=208034

amistre64 · Answer

thats not really a question ... its just stated information about the situation.

anonymous · Answer

sorry.

anonymous · Answer

1) Determine the acceleration of the system

2) Determine the tension T in the rope

those are the questions; thanks

amistre64 · Answer

Well, we should know by now that Force = mass x acceleration.

We know the total mass, and the force in the direction of motion is a scalar of the original force and the angle at which it is applied

right?

anonymous · Answer

yes

amistre64 · Answer

Force in the direction of motion:  89 cos(22) N
mass: (14+22)  kg

F = ma;  therefore  a = F/m

amistre64 · Answer

but, there might be something im leaving off, since it gives the Friction of kinetic, we might want the Force as a product of the kinetic coeff

amistre64 · Answer

$$F_k=\mu _k*\vec n$$

anonymous · Answer

i was thinking perhaps subtract the kinetic force from the pulling force and divide by m. but this did not work either unless I am calculating Fk wrong.

amistre64 · Answer

$$\vec n=mg$$hmm|dw:1348694530794:dw|