Block B in the figure below weighs 713 N. The coefficient of static friction between block and table is 0.20. Assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.
5 answers
Where is the knot? where is block A
the knot is inbetween block b and the wall. block a hangs from the knot. the knot is connected to the wall by a rope attached to the wall at 30 degrees from the horizontal
OK, break the problem into two parts, the rope tension into vertical and horizontal parts.
Since the wall and block both support the vertical equally, the vertical down force on the blockB is 1/2 weightA.
So forcefrictionB= mu*forcenormal
=mu(massB*g+1/2 massA*g)
now, that force of friction is opposing the horizontal part of tension.
horizontal: Tension*cosTheta, but
tension/.5Ma*g=sinTheta
or tension= .5Ma*g*sinTheta
horiztonal=friction force
.5Ma*g*sinTheta*cosTheta=mu*(MassB*g+.5Ma*g)
solve for MassA
Since the wall and block both support the vertical equally, the vertical down force on the blockB is 1/2 weightA.
So forcefrictionB= mu*forcenormal
=mu(massB*g+1/2 massA*g)
now, that force of friction is opposing the horizontal part of tension.
horizontal: Tension*cosTheta, but
tension/.5Ma*g=sinTheta
or tension= .5Ma*g*sinTheta
horiztonal=friction force
.5Ma*g*sinTheta*cosTheta=mu*(MassB*g+.5Ma*g)
solve for MassA
is Ma the mass of block a?
i still couldn't get the answer either