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.
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
is Ma the mass of block a?
i still couldn't get the answer either
To find the maximum weight of block A for which the system will be stationary, we need to consider the forces acting on the blocks.
First, let's identify the forces acting on each block:
For block B:
- Weight (Wb) = 713 N
- Normal force (Nb) = equal in magnitude and opposite in direction to the weight because block B is on a horizontal table.
For block A:
- Weight (Wa) = unknown (what we need to find)
- Tension force in the cord (T) = equal in magnitude and opposite in direction to the weight because the cord is assumed to be horizontal.
- Friction force between block A and the table (Fa) = μs * Nb, where μs is the coefficient of static friction and Nb is the normal force on block A.
Since the system is stationary, the net force on each block in the horizontal direction must be zero. This means that the friction force on block A must balance out the tension force in the cord.
Therefore, we can set up the following equation:
Fa = T
μs * Nb = Wa
μs * (Wa + Wb) = Wa
Now, substitute the given values:
0.20 * (Wa + 713 N) = Wa
0.20Wa + 142.6 N = Wa
142.6 N = 0.80Wa
Wa = 142.6 N / 0.80
Wa = 178.25 N
Therefore, the maximum weight of block A for which the system will be stationary is 178.25 N.