Home»Math Guides»3 Masses on inclined slope attached to pulley with friction physics problem
Three boxes attached on an inclined slope, with friction attached to pulley system
If you haven’t, review the 2-boxes on an inclined slope problem we solved previously here, it will go over the equations in more detail.
Consider the following diagram:
Example: Masses 1 and 2 are moving down the ramp. Find the tension between each of the ropes and the acceleration. Masses 1, 2 and 3 are 20 kg, 10 kg, and 5 kg, respectively. Theta is 30 degrees to the horizontal as illustrated in the diagram below. The coefficient of kinetic friction is 0.15.
Looking at the diagram, the inclined slope is at an angle of theta. We have three masses, labelled 1, 2 and 3. There are two separate tensions, one for the string connecting masses 1 and 2, and another tension for the string connecting masses 2 and 3. There will be friction opposite to the direction masses 1 and 2 are moving in.
You can make five different equations: 2 horizontal acceleration equations for 1 and 2, 2 vertical acceleration equations for 1 and 2, and a balance on the third mass.
In this problem, we said in the problem statement that the blocks are moving down the ramp, just to make the problem a little easier. That’s because you can really screw up the equations if you guess the system moves in the wrong direction and you end up with friction terms going in the wrong direction.
Let’s make equations for each of the masses. Remember that there is no acceleration in the vertical direction for masses 1 and 2 (the blocks don’t float off the ramp).
In general, we will end up with 3 equations and we will get 3 unknowns to solve for (tension 1, tension 2, and acceleration). The acceleration will be common in the system, all the masses react to each other together.
First mass:
\[{m_1}g\sin \left( \theta \right) – {T_1} – {F_{f1}} = {m_1}{a_x}\]
\[{F_{N1}} – {m_1}g\cos \left( \theta \right) = {m_1}{a_y} = 0\]
\[{F_{N1}} = {m_1}g\cos \left( \theta \right)\]
\[{F_f} = {\mu _k}{F_N}\]
\[{m_1}g\sin \left( \theta \right) – {T_1} – {\mu _k}{m_1}g\cos \left( \theta \right) = {m_1}{a_x}\]
Repeat for the second mass. It’s very similar, but now we have two tensions involved! The first tension acts in one direction, while the second tension acts in the opposite direction.
\[{m_2}g\sin \left( \theta \right) – {T_2} + {T_1} – {F_{f2}} = {m_2}{a_x}\]
\[{F_{N2}} – {m_2}g\cos \left( \theta \right) = {m_2}{a_y} = 0\]
\[{F_{N2}} = {m_2}g\cos \left( \theta \right)\]
\[{F_f} = {\mu _k}{F_N}\]
\[{m_2}g\sin \left( \theta \right) – {T_2} + {T_1} – {\mu _k}{m_2}g\cos \left( \theta \right) = {m_2}{a_x}\]
And the third mass, which will be easier to deal with:
\[{T_2} – {m_3}g = {m_3}{a_x}\]
Then all you need to do is plug in values and solve!
Our system of equations is:
\[{m_1}g\sin \left( \theta \right) – {T_1} – {\mu _k}{m_1}g\cos \left( \theta \right) = {m_1}{a_x}\]
\[{m_2}g\sin \left( \theta \right) – {T_2} + {T_1} – {\mu _k}{m_2}g\cos \left( \theta \right) = {m_2}{a_x}\]
\[{T_2} – {m_3}g = {m_3}{a_x}\]
Plugging in the values given in the question statement:
\[(20)(9.81)\sin ({30^o}) – {T_1} – (0.15)(20)(9.81)\cos ({30^o}) = (20){a_x}\]
\[(10)(9.81)\sin ({30^o}) – {T_2} + {T_1} – (0.15)(10)(9.81)\cos ({30^o}) = (10){a_x}\]
\[{T_2} – (5)(9.81) = (5){a_x}\]
Simplifying,
\[72.61 = 20{a_x} + {T_1} + 0{T_2}\]
\[36.3 = 10{a_x} – {T_1} + {T_2}\]
\[ – 49.05 = 5{a_x} + 0{T_1} – {T_2}\]
\[\left[ {\begin{array}{*{20}{c}}{20}&1&0\\{10}&{ – 1}&1\\5&0&{ – 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{a_x}}\\{{T_1}}\\{{T_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{72.61}\\{36.3}\\{ – 49.05}\end{array}} \right]\]
Solving,
Acceleration is 1.71 m/s2, the force of tension in the first rope is 38.4 N, and the force of tension in the second rope is 57.6 N.
Do you want to check to make sure your acceleration is in the correct direction? Look at the third mass’s equation:
\[{T_2} – {m_3}g = {m_3}{a_x}\]
T2 is 57.6 N, and -m3g is -5*9.81 = -49.05. That means that the force of tension is stronger than the force of gravity, meaning mass 3 rises, and consequently masses 1 and 2 going down the ramp to the left. So the directions are correct!
Let’s repeat the question, but now with everything moving in the opposite direction!
Example: Masses 1 and 2 are moving up the ramp. Find the tension between each of the ropes and the acceleration. Masses 1, 2 and 3 are 5 kg, 7 kg, and 80 kg, respectively. Theta is 30 degrees to the horizontal as illustrated in the diagram above. The coefficient of kinetic friction is 0.15.
Let’s reuse our equations and plug in the values, but, the term for friction is now in the opposite direction, so that sign is flipped.
\[(5)(9.81)\sin ({30^o}) – {T_1} + (0.15)(5)(9.81)\cos ({30^o}) = (5){a_x}\]
\[(7)(9.81)\sin ({30^o}) – {T_2} + {T_1} + (0.15)(7)(9.81)\cos ({30^o}) = (7){a_x}\]
\[{T_2} – (80)(9.81) = (80){a_x}\]
Simplify:
\[30.90 = 5{a_x} + {T_1} + 0{T_2}\]
\[43.26 = 7{a_x} – {T_1} + {T_2}\]
\[ – 784.8 = 80{a_x} + 0{T_1} – {T_2}\]
We cast the variables into a matrix and solve:
\[\left[ {\begin{array}{*{20}{c}}5&1&0\\7&{ – 1}&1\\{80}&0&{ – 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{a_x}}\\{{T_1}}\\{{T_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{30.90}\\{43.26}\\{ – 784.8}\end{array}} \right]\]
We end up with acceleration of -7.72 m/s2, force of tension 1 as 69.5 N, and force of tension 2 as 166.9 N.
Does the direction and sign of acceleration confuse you? Well to double check, look at the equation for the third block again:
\[{T_2} – {m_3}g = {m_3}{a_x}\]
We found tension 2 to be 166.9 N. -m3*g is -(80)(9.81) = 784.8 N. That means the force of gravity dominates for the third mass. If the third mass drops, then as a result masses 1 and 2 must travel up the ramp. Easy!
Try a physics problem with multiple pulleys involving the “conservation of string” concept