This is the problem and my work. I dont know where to go next. Please help!
N nerd New member Joined Jan 14, 2022 Messages 12 Feb 15, 2022 #1 This is the problem and my work. I dont know where to go next. Please help!
D Deleted member 4993 Guest Feb 15, 2022 #2 nerd said: This is the problem and my work. I dont know where to go next. Please help!View attachment 31129View attachment 31130 Click to expand... First, edited Define 'x' and 'y' (orthogonal) directions according to your convenience. Cut the rope and replace it with a tension T (directed along the rope) Draw the FREE BODY DIAGRAM of the tether ball (assumed to be a particle, with tension in the rope (T), weight W (= 1 lb) & the pulling force (u). Now apply conditions of equilibrium. ∑Fx=0\displaystyle \sum F_x = 0 ∑Fx=0 .... and .... ∑Fy=0\displaystyle \sum F_y = 0 ∑Fy=0 Continue ..... Last edited by a moderator: Feb 15, 2022
nerd said: This is the problem and my work. I dont know where to go next. Please help!View attachment 31129View attachment 31130 Click to expand... First, edited Define 'x' and 'y' (orthogonal) directions according to your convenience. Cut the rope and replace it with a tension T (directed along the rope) Draw the FREE BODY DIAGRAM of the tether ball (assumed to be a particle, with tension in the rope (T), weight W (= 1 lb) & the pulling force (u). Now apply conditions of equilibrium. ∑Fx=0\displaystyle \sum F_x = 0 ∑Fx=0 .... and .... ∑Fy=0\displaystyle \sum F_y = 0 ∑Fy=0 Continue .....
N nerd New member Joined Jan 14, 2022 Messages 12 Feb 15, 2022 #3 This is my work so far. I reached a dead-end, though. Am I doing anything wrong?
BigBeachBanana Senior Member Joined Nov 19, 2021 Messages 2,277 Feb 16, 2022 #4 Right triangles are your friends. To find the resulting tension ∣t→∣|\overrightarrow{t}|∣t∣: cos(θ)=∣w→∣∣t→∣\cos(\theta)=\frac{|\overrightarrow{w}|}{|\overrightarrow{t}|}cos(θ)=∣t∣∣w∣To find the magnitude of ∣u→∣|\overrightarrow{u}|∣u∣: tan(θ)=∣u→∣∣w→∣\tan(\theta)=\frac{|\overrightarrow{u}|}{|\overrightarrow{w}|}tan(θ)=∣w∣∣u∣
Right triangles are your friends. To find the resulting tension ∣t→∣|\overrightarrow{t}|∣t∣: cos(θ)=∣w→∣∣t→∣\cos(\theta)=\frac{|\overrightarrow{w}|}{|\overrightarrow{t}|}cos(θ)=∣t∣∣w∣To find the magnitude of ∣u→∣|\overrightarrow{u}|∣u∣: tan(θ)=∣u→∣∣w→∣\tan(\theta)=\frac{|\overrightarrow{u}|}{|\overrightarrow{w}|}tan(θ)=∣w∣∣u∣