- #1
GrimesA
- 20
- 0
So far I thought that i'd solve it like this:
Weight of block W = mg
Component of W parallel to slope = Wsinθ
Component of W perpendicular to slope = Wcosθ
Call R the normal reaction force of the slope on the block.
In the direction perpendicular to the slope, the force are balanced:
R = F + Wcosθ
The limiting frictional force, Fr is:
Fr = μR = μ(F + Wcosθ)
In the direction parallel to the slope, when on the point of slipping the limiting frictional force is just balanced by the component of W parallel to slope:
Fr = Wsinθ
μ(F + Wcosθ) = Wsinθ
μF + μWcosθ) = Wsinθ
μF = Wsinθ - μWcosθ)
. . = W(sinθ - μcosθ)
F = W(sinθ - μcosθ)/μ
. = mg(sinθ – μcosθ)/μ
However, when I plug in all of my numbers, I get -215N. And when I put that into my homework, it is wrong. Is this a computational error, was my formulas flawed?