still ba ng the y direction. The force of gravity in the x direction (mg) sint6)
ID: 1874741 • Letter: S
Question
still ba ng the y direction. The force of gravity in the x direction (mg) sint6) must be balanced by the force of static friation to be parallel to the inclined surface and the y Equation (1)2 F.-mg sin()-h-o Equation (2) F,-"-mg cos()-0 Equation (3), _ mgsin()-(cos(,) sinh-stand Equation (4) -tan() Similarly, along the y direction, the normal force ii from the surface must balance the corresponding component of force from gravity Solving Equation (2) for mg-n(cos()-1 and substituting into Equation (1) gives sin(8) ntan() when the block is on the verge of slipping, -e, and the force of static friction has reached its maximum valueh-".A+ ande into tuition 00pes For the measured critical angle of -26.5°, we find from Equation (4) that A, tan 26.50 Finalize Once the block starts to move at > it accelerates down the inine and the free of friction ish- n. if is reduced to a value less than , however, K may be possible to find an angle e such that the block moves down the incline with constant speda case, Equations (1) and (2) with f, replaced by fi lead to an expression that gives the value of the kinetic fiction where e, it is possible to find an angle 0, such that the block moves down the Incline constant speed as a particle in equilibrium again (a, - 0). In this case, use Equations (1) and (2) with f replaced byto find .-tance, where as a particle in equilibrium again (with acceleration a 0). In this coefficient of Need Help? Talk to a Tutor Show My Work (Optional, @Explanation / Answer
The answer is just plugging in the value of critical angle in the equation and evaluating the tangent in degrees..
= tanc
=tan(26.5)
=0.4986
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