Answer:
n = 1/5 and m = 3/5
Explanation:
The given quantity is :
Where
The dimension of [A] = [LT]
The dimension of [B] = [L²T⁻¹]
The dimension of [C] = [LT²]
We need to find the dimensions of n and m values.
Using dimensional analysis,
![[LT]=[L^2T^{-1}]^n[LT^2]^m\\\\\ [LT]=L^{2n}T^{-n}\times L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}](https://tex.z-dn.net/?f=%5BLT%5D%3D%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%5D%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%7DT%5E%7B-n%7D%5Ctimes%20L%5EmT%5E%7B2m%7D%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%2Bm%7DT%5E%7B2m-n%7D)
Comparing both sides,
2n+m=1 ....(1)
-n+2m=1 ,.....(2)
Solving (1) and (2), we get :
n = 1/5 and m = 3/5
Hence, this is the required solution.
By equation of equilibrium and friction:
Fb = Kx = 15(0.175) = 2.625 kN.
The wedge is on the verge of moving right then slipping will
have to occur at both contact surfaces.
Fa = usNa = 0.35Na
Fb = 0.35Nb
Nb = 2.625 = 0; Nb = 2.625 kN
Nacos10 – 0.35Na sin 10 = 2.625 = 0
Na = 2.841 kN
P – (0.35 * 2.625) – 0.35 (2.841) cos 10 – 2.841 sin 10 = 0
P = 2.39 kN
1.) temperature change, 2.) color change, 3.) gas formation
