4. Given the area of the left piston (under the 20-‐kg mass) is 4 m2, what is the area of the right piston (under the 600-‐kg mass)?

5. If the mass on top of the left piston was doubled to 40 kg, how would this affect the upward force on the right piston? Use the areas given and calculated in #4.

Question

4. Given the area of the left piston (under the 20-­‐kg mass) is 4 m2, what is the area of the right piston (under the 600-­‐kg mass)?

5. If the mass on top of the left piston was doubled to 40 kg, how would this affect the upward force on the right piston? Use the areas given and calculated in #4.

anonymous · Answer

Assuming the diagram is of a U-tube type piston configuration and the fluid between the two pistons of incompressible, this is simply a balance of pressures. Since pressure is defined as$$P = {F \over A}$$and the pressure on the left and right pistons must be the same, we can infer that$${F_l \over A_l} = {F_r \over A_r} \rightarrow {m_l g \over A_l} = {m_r g \over A_r} \rightarrow {m_l \over A_l} = {m_r \over A_r}$$This expression can be manipulated to solve for which ever quantity you desire.

anonymous · Answer

This is the piston I'm working with for these question.

anonymous · Answer

after working out the two questions using your equation, the answer i got for #4 was 120 being the area under the right piston and #5 i think the upward force will be halved still not sure if that is correct. pls let me know im right

anonymous · Answer

120 is correct for #4. 

For number 5, if we double the mass, we will double the force exerted by the left piston because gravitational acceleration is constant. (Note this is not always the case, i.e. if we were had external forces acting on the pistons.) Since the area is constant, we can see that pressure is doubled. $$F_l = P_l * A_l ~ {\rm and}~ 2*F_l = 2*P_l * A_l$$The pressure on the left side must equal the pressure on the right side. $$P_l = P_r$$Therefore, if the pressure on the right is double, the pressure on the left must be doubled. As seen by the first equation, the force must be doubled on the left side. Note that for the pistons to remain static (no motion) the downward force exerted by the piston equals the upward force exerted by the hydraulic fluid. Therefore, if we double the mass of the right, the mass on the left will move upward.

anonymous · Answer

we are doubling the mass on the left, the 20kg not the right.

anonymous · Answer

to maintain equilibrium, we have to balance the pressure.
P=F/A
P ON LEFT = 20/4=5
P ON RIGHT=600/x=5
HENCE x =120m^2 IN #4
if the mass on the left is doubled, the area on left will be halved i.e. 60m^2
this is true for incompressible fluids only