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?
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.
To find the answers to these questions, we can start by analyzing the forces acting on the mass on the frictionless wedge.
(a) To find the angle of inclination, θ, of the wedge, we need to consider the forces acting on the mass. The two main forces are the weight of the mass (mg), acting vertically downward, and the normal force (N), acting perpendicular to the inclined surface.
Since the mass remains stationary relative to the wedge, the net force acting on the mass in the horizontal direction must be zero. This means that the horizontal component of the weight force (mg*sinθ) must be balanced by the horizontal component of the normal force (N*sinθ).
Since the wedge has a horizontal acceleration of 15 m/s^2 to the right, we can write Newton's second law in the horizontal direction:
m*a = F_net = N*sinθ - mg*sinθ,
where m is the mass of the object (2.0 kg) and a is the acceleration of the wedge (15 m/s^2).
Since the mass remains stationary relative to the wedge, its acceleration is zero. Therefore, we have:
0 = N*sinθ - mg*sinθ.
Simplifying this equation, we can factor out the sinθ:
0 = (N - mg)*sinθ.
For the mass to remain stationary, the sine of θ must be zero (sinθ = 0). This implies that the angle of inclination, θ, is 0°.
(b) The normal force, N, is the force exerted by the inclined surface perpendicular to the surface. To find the magnitude of the normal force, we can use the equation we derived in part (a):
0 = N - mg.
Solving for N, we have:
N = mg.
Substituting the values, where m = 2.0 kg and g is the acceleration due to gravity (approximately 9.8 m/s^2), we get:
N = 2.0 kg * 9.8 m/s^2 = 19.6 N.
Therefore, the magnitude of the normal force exerted on the mass by the incline is 19.6 N.
(c) If the wedge were given a greater acceleration, the mass would no longer remain stationary relative to the wedge. To analyze what would happen, we need to consider the forces acting on the mass again.
With a greater acceleration, the net horizontal force on the mass would no longer be zero. This would cause the mass to be accelerated horizontally in the same direction as the wedge.
If the wedge's acceleration is increased further, the mass would accelerate faster in the same direction. The angle of inclination (θ) would still be 0°, but the magnitude of the normal force exerted on the mass would decrease due to the larger horizontal acceleration. The new magnitude of the normal force could be determined using the same equation, N = mg - ma, where a is the increased acceleration.