How Do Newton's Laws Apply to a Lemon Half on a Frictionless Surface?

AI Thread Summary
The discussion revolves around calculating the third force acting on a lemon half on a frictionless surface under different conditions of motion. For zero velocity, the required force components were found to be x = -2.11 and y = 1.11. When considering constant velocity, the third force components were adjusted to x = -2.01 and y = 1.12. For varying velocity, the acceleration was derived from the velocity function, leading to forces of x = 2.27 and y = -0.84. The participant expresses confusion over the calculations and seeks clarification on the application of Newton's laws in each scenario.
iamkristing
Messages
33
Reaction score
0
1. A figure shows an overhead view of a lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 4.70 N (in the negative x and positive y direction) and is at θ 1 = 29°. Force F2 has a magnitude of 7.00 N (in the positive x and negative y direction) and is at θ 2 = 29°. The lemon half has mass 0.0200 kg.

(a) What is the third force if the lemon half has zero velocity?


(b) What is the third force if the lemon half has constant velocity? v = (13.0 - 14.0 ) m/s


(c) What is the third force if the lemon half has a varying velocity? v = (13.0t - 14.0t ) m/s, where t is time in seconds


2. F= m*a


3. I divided the two forces into x and y components. I then added them together and found the x and y component needed for the lemon not to move. I'm not sure how to use these components to find velocity though.

Also, is there a difference between constant and varying velocity for this problem?
 
Physics news on Phys.org
You need to show some working or an attempt before someone helps you.
 
sorry I have attempted.

I found the x and y components of the first force to be x= -4.11 and y=2.28 and the second force to be y= -3.39 and x=6.12

for the lemon not to move, I found that the third force's x and y components were equal to x=-2.11 and y=1.11

now I did the same except when solving for the third force i set each component equal to the constant velocity. I got x=10.99 and y=-12.89

for the varying velocity question, I know the derivative of velocity is acceleration. So i found the derivative of that function and then applied F=ma. I got x=0.26 abd y=-0.28

Now none of these answers are right and I can't see where I'm going wrong...
 
I just tried the problem again, taking into account the velocity and acceleration. I thought I missed that before.

So I used the sum of the forces = mass*acceleration

I thought, with zero velocity, no other force would be needed because the velocity doesn't change.

Also, for constant velocity, I found the third force vector needed. I found that to be x=-2.01 and y=1.12

Now for varying velocity, I took the derivative of the velocity the problem gave me. I found the acceleration to be 13 in the x direction and 14 in the y direction. I used F=ma and plugged in the mass of the lemon and the acceleration and got:

F in x direction = 2.27
F in y direction = -0.84

Still though, I cannot come up with the answer. Any help or point in the right direction will be helpful!
 
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