Forces in the triangle and pulley system

AI Thread Summary
Two blocks connected by a string, one on a 35-degree incline and the other hanging, create a system where the acceleration of the hanging block must be determined. The equations of motion depend on the chosen coordinate system, where forces acting in the positive direction are considered positive. For the hanging block, the net force is represented as F(net) = mg - T, indicating that tension is subtracted from the weight due to its upward direction. The incline's angle affects the forces acting on the block on the incline, and careful consideration of the coordinate system is crucial for accurate calculations. Understanding the relationship between tension and gravitational force is key to solving for the acceleration of the system.
godkills
Messages
42
Reaction score
0
Two blocks are connected by a string-one on an incline, the rope draped over a pulley, and one hanging from the string. The smooth inclined surface makes an angle of 35 with the horizontal, and the block on the incline has a mass of 5.7 kg. The mass of the hanging block is 3.2kg. Find the direction and magnitude of the hanging block's acceleration.

Can someone explain to me why the smaller block should be represented as this equation?

For m: F(net) = ma = mg - T.

and not T - mg = ma? like the first one
 
Physics news on Phys.org
Hi,

To fix the problem you are encountering, draw a diagram and set up the correct coordinate axis.

Untitled-10.png


**Woops that angle should be 35 degrees, but you get the picture :)
 
godkills said:
Can someone explain to me why the smaller block should be represented as this equation?

For m: F(net) = ma = mg - T.

and not T - mg = ma? like the first one

These equations are relative to the coordinate axis chosen. Essentially you have two choices in this problem: the system accelerating right, and the system accelerating left.

For example, let's choose 'acceleration of the system to the left' to be positive.

This means that all forces pointing left will be positive. We can now set up a system of equations for each box.

For the 5.7kg box on the incline, there are two forces acting on it that will cause the objects to go up or down the incline; since we chose 'accelerating left' as positive, all forces that will cause the box to slide down the incline, will be positive.

The two forces we have are: Fgx and FT
Fgx points down the incline so we make it positive (since we chose 'to the left' as positive').

FT points up the incline, and thus is assigned a negative value.

Therefore, our Fnet equation for the 5.7kg box will be: Fgx+(-FT)=ma

Apply this to the other box, then add the equations together and solve for a (the FT will cancel)
 
yes i had the FBD drawn and i understand i think how the larger mass should be moving the the left due to it's heavy weight but i am not getting why for the second mass tension is subtracted from the weight.

i drew a triangle perpendicular to the larger mass. indicating sin(35)mg is positive.
 
Last edited:
godkills said:
yes i had the FBD drawn and i understand i think how the larger mass should be moving the the left due to it's heavy weight but i am not getting why for the second mass tension is subtracted from the weight.

i drew a triangle perpendicular to the larger mass. indicating sin(35)mg is positive.

Careful, just because the mass is greater doesn't mean it is going to accelerate that way (the angle affects Fgx)

For example if the angle was 1o and there was no friction, the box would move right (think, there is no way it could move left)

I picked accelerating left as an arbitrary set up--you could have said 'to the right' is positive; however, the coordinates would be reversed.
 
godkills said:
i am not getting why for the second mass tension is subtracted from the weight.

Tension in the smaller mass is acting upwards, which would cause the system to move left, and therefore would be positive (from the axis we set up). Think, pulling on the rope holding the smaller mass will cause it to accelerate which way?

Untitled-12.png


**To the right will also be the same as saying the system accelerates towards the smaller mass

** To the left will also be the same as saying the system accelerates towards the larger mass.
 
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