A 2.0 kg mass rests on a frictionless wedge that has an acceleration of 15 m/s2 to the right. The mass remains stationary relative to the wedge, moving neither higher nor lower. (a) What is the angle of inclination, θ, of the wedge, i.e., the angle between the inclined surface and the horizontal? (b) What is the magnitude of the normal force exerted on the mass by the incline? (c) What would happen if the wedge were given a greater acceleration?
2 answers
not my prob
the horizontal component of acceleration down the wedge is 15m/s. So down the wedge, the component is 15CosTheta. The down the incline component of weight is mg*sinTheta.
So, since it is not moving relative to wedge, mgSinTheta=15cosTheta
Tan Theta=15/mg
b. normal force? consists of weight mgCosTheta plus 15SinTheta
c. If the 15 becomes bigger, then it slides upward.
So, since it is not moving relative to wedge, mgSinTheta=15cosTheta
Tan Theta=15/mg
b. normal force? consists of weight mgCosTheta plus 15SinTheta
c. If the 15 becomes bigger, then it slides upward.
3 answers