How Does the Normal Force Affect Dynamics in Different Frames of Reference?

[radagast_]

Homework Statement

here:
http://img144.imageshack.us/img144/7518/26700576ok1.gif

Homework Equations

The Attempt at a Solution

well, I tried to solve the motion equation from the inside table. I think there should be a d'elambertian force to the left, and friction force to the right, so:
Fd - miu*N = m1*a . Now, I am not sure about the Normal force- need it be N=m1+m2, or just m1 (I think it's the first option).
Then my main problem - about the d'elambertian force: Is the Fd equal to the external frame's F, or are the accelerations equal, and the forces need to be calculated via the acceleration and masses?
And finally: when I try to calculate on the external frame (the cart): need the motion equation be equaled to just Ma, or all the masses, ie (m1+m2+M)a ?

Thanks in advance..

 

[radagast_]

anyone?

 

[HallsofIvy]

Actually, there are two different situations to think about. If the force is applied to the cart at t= 0, then, since there is friction between the cart and the table will start moving as well as the cart. Since there is NO friction between the table and m2, m2 will NOT start moving. Eventually, it will just slide off the table and fall onto the floor of the cart. But while it is on the table, it contributes to the normal force the table makes with the floor of the cart: the normal force is (m1+ m2)g so the friction force exerted by the table on the cart is $\mu(m1+ m2)g= 0.8(1+ 9)(9.81)= 8(9.81)= 78.48 N. The Net force 1120- 78.48= 1041.52 N. Since the total mass of the cart is M0+ m1= 81 kg (you do NOT count m2 since it is not being accelerated) The carts acceleration is 1041.52/81= 12.8 m/s².<br /> <br /> The other situation is after m1 has slid off the table. It no longer contributes to the normal force increasing the friction force of the table on the floor of the cart but might contribute to the total mass if there is friction between it and the floor of the cart.$

 

[radagast_]

Thanks for the thorough response!

 

[radagast_]

Actually, there are two different situations to think about. If the force is applied to the cart at t= 0, then, since there is friction between the cart and the table will start moving as well as the cart. Since there is NO friction between the table and m2, m2 will NOT start moving. Eventually, it will just slide off the table and fall onto the floor of the cart. But while it is on the table, it contributes to the normal force the table makes with the floor of the cart: the normal force is (m1+ m2)g so the friction force exerted by the table on the cart is $\mu(m1+ m2)g= 0.8(1+ 9)(9.81)= 8(9.81)= 78.48 N. The Net force 1120- 78.48= 1041.52 N. Since the total mass of the cart is M0+ m1= 81 kg (you do NOT count m2 since it is not being accelerated) The carts acceleration is 1041.52/81= 12.8 m/s².<br /> <br /> The other situation is after m1 has slid off the table. It no longer contributes to the normal force increasing the friction force of the table on the floor of the cart but might contribute to the total mass if there is friction between it and the floor of the cart.$

$<br /> You said to include both M0 and m1 as the objects being accelerated. But, aren't they moving at different speeds? so how can I do F=(m0+m1)a, whilst m0 and m1 aren't moving together?$