Magnitude and Acceleration Problem

Click For Summary
Three astronauts are using jet backpacks to maneuver a 148 kg asteroid, applying forces of 29 N, 54 N, and 40 N at specified angles. The problem involves calculating the asteroid's acceleration magnitude and angle using the formula F=ma. Initial calculations for the x and y components of acceleration were presented, but an error was identified in the y-component due to the incorrect sign of the force acting downward. After correcting this, the final acceleration magnitude is calculated to be approximately 0.748 m/s², with an angle of 26.5 degrees relative to the positive x-axis. The discussion emphasizes the importance of accurately accounting for direction in vector components.
MFlood7356
Messages
37
Reaction score
0
1. Three astronauts, propelled by jet backpacks, push and guide a 148 kg asteroid toward a processing dock, exerting the forces shown in the figure, with F1 = 29 N, F2 = 54 N, F3 = 40 N, θ1 = 30°, and θ3 = 60°. What is the (a) magnitude and (b) angle (measured relative to the positive direction of the x axis) of the asteroid's acceleration?

2. F=ma

3. So I've been trying to do this problem and I can't seem to figure it out. I'm on my last attempt on my online homework so I really need some help. Here's how far I've gotten:
Ax=((29cos30)+(54cos0)+(40cos60))/148=0.667
Ay=((29sin30)+(54sin0)+(40sin60))/148=0.332
sqrt0.3322+0.6672=0.748
invertan(0.332/0.667)=26.5degrees
 
Physics news on Phys.org
Whether angle θ3 =60 degrees is above or below the + x-axis?
 
The third angle is below the x axis.
 
MFlood7356 said:
The third angle is below the x axis.
In that case
Ay=((29sin30)+(54sin0)+(40sin60))/148=0.332
this equation is wrong. It should be
Ay=((29sin30)+(54sin0) - (40sin60))/148=0.332
 
So the only thing I did wrong was not negate that part?
 
Yes.
 
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
5
Views
8K
  • · Replies 22 ·
Replies
22
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
4
Views
4K