Frictionless plane at angle of theta=69degrees

Click For Summary
The discussion revolves around calculating the acceleration of block 1 on a frictionless inclined plane tilted at 69° with respect to the vertical. The masses involved are 6 kg for block 1 and 3.9 kg for block 2, which is hanging from a pulley. The user attempts to apply Newton's second law (F=ma) but finds the problem complex, particularly in defining the forces acting on the blocks and the role of the pulley. Clarifications are requested regarding the definitions of tension (T) and weight (W), as well as the forces acting on block 1. The conversation highlights the challenges in visualizing the system and setting up the equations correctly.
mrshappy0
Messages
97
Reaction score
0

Homework Statement



A table is tilted at an angle of θ = 69° with respect to the vertical. Find the magnitude of the new acceleration of block 1. Mass of the block 1 which is on the incline of the plan is 6. Mass of block 2 which is hanging freely from the pulley is 3.9.

Homework Equations



F=ma

The Attempt at a Solution



I drew the free body diagram and started working on block one and set the x-axis parallel to the plane and have the y-axis perpendicular to make it easier. Fnet,alongx=T-Wtan(theta)=m1(a). Then I went on to work on Fnet,alongy=.. and it started looking really complicated. Not sure if I am approaching it correctly.
 
Physics news on Phys.org
I am struggling to grasp the problem definition. I can understand that there is a table, which makes an angle of 69^{\circ} with the vertical. It initially as a block M_{1} on the table at rest, and you wish to know the net acceleration. I do not understand your reference to a pulley and block 2, where is the pulley?

Could you also expand on what T and W are?

From what I can see you should have 2 forces acting on the mass, M1. So for your Fnet_along y there should be a maximum of two terms?

Thanks
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
9
Views
5K
  • · Replies 68 ·
3
Replies
68
Views
13K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K