How to Calculate Mass of a Block Using Force Equations?

  • Thread starter Thread starter miktalmyers
  • Start date Start date
  • Tags Tags
    Block Mass
AI Thread Summary
The discussion focuses on calculating the mass of a block using force equations. The equations provided include the weight of the block and the forces acting on it, including friction and tension. It is clarified that the force FA represents the weight of block mA, and it must be less than mA's weight for downward acceleration. The calculations for friction forces (FfB and FfC) and the resulting tension force (FB) are detailed, leading to the conclusion that mA is approximately 794 grams. The importance of correctly applying Newton's second law in this context is emphasized.
miktalmyers
Messages
2
Reaction score
0
Calculating mass of a block...

Homework Statement



Physicsquestions1.jpg


Homework Equations



Force equations/Kinetic equations


The Attempt at a Solution



FA = mA * 9.8 m/sec2

FA = FB
where FB is the force acting through the string from A to B

The forces on B are friction from the table, friction from C, and the force it takes to move B's mass at 0.8 m/sec2
FB = FfB + FfC + (1.4 kg * 0.8 m/sec2)

FfC = 0.3 * ( 0.6 kg * 9.8 m/sec2 )

FfC = 1.764 N

FfB = 0.25 * [ (1.4 kg + 0.6 kg) * 9.8m/sec2 ]

FfB = 4.9 N

FB = 1.764 N + 4.9 N + (1.4 kg * 0.8 m/sec2)

FB = 7.784 N

FA = 7.784 N

7.784 N = mA * 9.8 m/sec2

mA = 0.794286 kg ~ 794 grams
 
Physics news on Phys.org


No one?
 


miktalmyers said:

Homework Statement



Physicsquestions1.jpg


Homework Equations



Force equations/Kinetic equations


The Attempt at a Solution



FA = mA * 9.8 m/sec2
This is not FA (the string tension), this is the weight of block mA. FA must be less than the weight of block mA, or else it can't accelerate downward.
FA = FB
where FB is the force acting through the string from A to B
yes, this is correct.
The forces on B are friction from the table, friction from C, and the force it takes to move B's mass at 0.8 m/sec2
stay away from pseudo forces, the horizontal forces on B are friction from the table, friction from C, and the tension force FB
FB = FfB + FfC + (1.4 kg * 0.8 m/sec2)
yes...
FfC = 0.3 * ( 0.6 kg * 9.8 m/sec2 )

FfC = 1.764 N

FfB = 0.25 * [ (1.4 kg + 0.6 kg) * 9.8m/sec2 ]

FfB = 4.9 N

FB = 1.764 N + 4.9 N + (1.4 kg * 0.8 m/sec2)

FB = 7.784 N

FA = 7.784 N
I'm not checking your numbers, but your equations are OK
7.784 N = mA * 9.8 m/sec2

mA = 0.794286 kg ~ 794 grams
This is your error as i noted above; the weight mA acts down, and FA acts up, solve for mA using Newton 2.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top