Finding angle for specific moment

[werson tan]

Homework Statement

i have question with part B
my working is
13.2 cos tetha (0.86x10^-3)+ 13.2 sin tetha (1.22x10^-3) = 1.95
i gt stucked here. how to proceed? i only have one equation , how to solve two unknown ?

Homework Equations

The Attempt at a Solution

 

[andrevdh]

You should not assume the moment is 1.95 for part A. From the given information you should be able to see that theta is 30^o, which enables you to calculate the moment of P about A for question a.

 

[werson tan]

You should not assume the moment is 1.95 for part A. From the given information you should be able to see that theta is 30^o, which enables you to calculate the moment of P about A for question a.

i have done part A , i just have problem with part b , i did not assume the moment is 1.95 for part A. i gt -1.788N for part a

 

[andrevdh]

Another way is to use the formula τ = P r sin(θ) where θ is the angle between P and r. You need to make a drawing to figure our what the relationship between θ and α is then.

 

[werson tan]

Another way is to use the formula τ = P r sin(θ) where θ is the angle between P and r. You need to make a drawing to figure our what the relationship between θ and α is then.

(13.2cos (alpha) ) ( surd(0.086² +0.122² ) )= 1.95 , alpha = 8.23 degree , still wrong !

 

[andrevdh]

Did you take the square root to obtain r? The angle in the equation is not necessarily alpha. Make a drawing to discover the relationship.

 

[werson tan]

Did you take the square root to obtain r? The angle in the equation is not necessarily alpha. Make a drawing to discover the relationship.

my working is 13.2 (1-(sin theta)^2) )^(0.5) (0.86x10^-3) + 13.2(sin theta)(122x10^-3) = 1.95 , so i gt the theta = 56.42 , but the ans given is 50.6 degree and 59.1 degree

 

[andrevdh]

I seem to get 47.5^o for α

 

[haruspex]

i gt stucked here. how to proceed? i only have one equation , how to solve two unknown ?

You do not have two unknowns. You have two functions, cos and sin, of one unknown.
Do you know the formula for sin(a+b)? Can you see how to apply it in reverse to an expression like $A\cos(\theta)+B\sin(\theta)$?

 

[haruspex]

I seem to get 47.5^o for α

Unfortunately the arc sin function gets very steep near 1, so it's hard to get enough precision. I get more like 46.6 or 63 degrees.

 

[werson tan]

You do not have two unknowns. You have two functions, cos and sin, of one unknown.
Do you know the formula for sin(a+b)? Can you see how to apply it in reverse to an expression like $A\cos(\theta)+B\sin(\theta)$?

can i do in this way ?
13.2 (1-(sin alpha)^2) )^(0.5) (0.86x10^-3) + 13.2(sin alpha)(122x10^-3) = 1.95 , so i gt the alpha = 56.42 , but the ans given is 50.6 degree and 59.1 degree

 

[haruspex]

can i do in this way ?
13.2 (1-(sin alpha)^2) )^(0.5) (0.86x10^-3) + 13.2(sin alpha)(122x10^-3) = 1.95 , so i gt the alpha = 56.42 , but the ans given is 50.6 degree and 59.1 degree

All those numbers make it too hard to follow what you are doing. Use symbols and show your working to get that equation.

 

[werson tan]

All those numbers make it too hard to follow what you are doing. Use symbols and show your working to get that equation.

sorry , pls ignore my previous working. i have changed the cos aplha = (1- (sin aplha )^2 )^0.5 , but i still coulnt get the ans , since the sin alpha value is greater than 1

 

[haruspex]

sorry , pls ignore my previous working. i have changed the cos aplha = (1- (sin aplha )^2 )^0.5 , but i still coulnt get the ans , since the sin alpha value is greater than 1

$\sqrt{1-\sin^2(\alpha)}$ is not $1-\sin(\alpha)$.