248.7N was the correct answer...so thanks!
I guess no matter what the angle T is at it will always be equal to the tension on the other side of the pulley
No don't say sorry, I appreciate your help!
I only have two tries left so I might try to work it out again...maybe i'll get some more understanding about this question
I don't have the answer sheet, but its an online assignment and when I enter an answer It tells me if its wrong.
The only thing I can figure is that the tension on the left side of the larger pulley is different than the tension on the right side of that pulley
I got an answer of T = 351.7N but unfortunately that's not right.I came up with the statement:
T1 + T2 - M2 - MC = 0
Where T1 and T2 = Tsin45
So:
2Tsin45 = 497.367N
and T = 351.7N
But this isn't the right answer... Thanks again for your help!
Ok that seems like it makes sense...one more question though.. how would the pulley with mass = m1 affect the system? Since it's attached to the ceiling wouldn't it's weight be canceled out by T3?
Homework Statement
A crate is pulled up using frictionless pulleys in the manner shown in the figure. The angle is 45 degrees. The masses are, for the small pulley, m1=3.5 kg, for the traveling pulley, M2=6.5 kg, and for the crate, MC=44.2 kg. What is the minimum tension with which the...
Homework Statement
An example of a precipitation reaction: Aqueous silver nitrate (AgNO3) is added to a solution containing potassium chloride (KCl) and the precipitation of a white solid, silver chloride is observed. (Zumdahl, 2005)
The silver chloride (AgCl) has formed a solid, which is...
a) Finding the horizontal distance after 0.25 s is fairly straightforward:
Given: vx = 1.45 m/s
dx = vxt
= (1.45 m/s)(0.25 s) = 0.36 m
Finding the vertical height at 0.25 s:
Given: v1y = 0
dy = 0 + 0.5a(t)^2
= 0.5(-9.8)(0.25)^2
= -0.31 m
1.3 m - 0.31 m = 0.99 m
b) ... same process
c)...
Sorry I can't be of much more help. I'm only doing entry level physics myself (conveniently, I was doing some torque problems today), but I've never done any torque problems like this.
Truthfully, I'm really confused by these parts:
I'm just assuming that's the torque of the rope? But how is...
You used sin 33 as the angle for force gravity between the person's legs and the horizontal.
This angle is meant to be between the force gravity (which is always straight down) and the person's legs, so instead of using 33, you use 90 - 33 = 57.
Try using:
746.76 N *sin(57)*1.2
For the...
I find the question strange, because really, there is two pivot points. One where the rope is attached to the cliff and also at the feet of the person who is 33 degrees above the horizontal.
How are you deciding whether the force of gravity on the person is clockwise or counterclockwise? And...