Force of Friction w/o Coefficient of Friction Given

AI Thread Summary
To determine the force of friction acting on block M1, the user is trying to calculate the tension in the string and the net force. They have already calculated the acceleration as 4.202 m/s² and are using the formula T = M2(-A) + M2(9.8) to find the tension force. However, they are confused about how to derive the net force and tension correctly. Additional guidance is provided to set up a force equation for block M1 to facilitate the calculations. The discussion emphasizes the importance of clearly distinguishing between tension and time in the calculations.
stehat
Messages
1
Reaction score
0
I am having trouble determining what the force of friction of the box on a table described as below:

Two blocks are arranged at the ends of a massless string. M1 is on the table. M2 is attached to another side of the rope hanging off of the side of a table with a massless pulley (90 degrees from the table).

M1= 4.78KG
M2 = 3.26 KG

When M2 has fallen through .396m, its downward speed is 1.29 m/s. Acceleration of gravity is 9.8. What is the frictional force between M1 and the table?

My first step was to find the acceleration with I figured to be 4.202. (By taking the acceleration formula A=V2-V1/T2-T1. I figured the time in this formula by using the velocity formula and solved for T.

My next step was to figure out the tension force which I think is where I am getting confused on where to go. I want to find the Tension force and subtract that from the Net force to find the force of friction, but I am having difficulty determining the net force an now I am not so sure I am figuring out the tension force correctly.

I am calculating the Tension force by taking T=M2(-A)+M2(9.8). Can you please help me finish?
 
Physics news on Phys.org
welcome to pf!

hi stehat! welcome to pf! :smile:

(btw, always use a small "t" for time, so as not to confuse with tension :wink:)
stehat said:
My first step was to find the acceleration with I figured to be 4.202. (By taking the acceleration formula A=V2-V1/T2-T1. I figured the time in this formula by using the velocity formula and solved for T.

i don't follow this: what formula did you use? :confused:
I am calculating the Tension force by taking T=M2(-A)+M2(9.8).

yes, M2a = M2g - T …

now do a similar F = Ma equation for the other block :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top