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?
draw the figure. there are two forces operating on the mass, mg downward, and a pushing force horizontal. Resolve both of those into components up /down the wedge surface.
mg resolves down the wedge as mgSinTheta downward
the pushing force has an equal and opposite force (inertia) from the mass of ma, which has an upward componsnt of ma CosTheta
so if it is not moving up or down, then these are equal.
ma CosTheta=mgSinTheta solve for theta.
Normal force:
mg normal force is mg CosTheta
ma normal force is maSinTheta
total normal force is the sum.
that didn't help at all... sorry
Thank You! it took me a while but that was all right and it worked. Thank You bobpursley sooo much!
To find the answers to these questions, we can analyze the forces acting on the system.
(a) To find the angle of inclination, θ, we need to consider the forces acting on the mass. Since the mass remains stationary relative to the wedge, we know that the net force acting on it must be zero. The forces involved are the gravitational force (mg) and the normal force (N) exerted by the incline.
Let's break down the forces:
- The gravitational force acting vertically downwards is given by mg, where m is the mass (2.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The normal force N is perpendicular to the incline.
The net force along the horizontal direction is provided by the acceleration of the wedge, which is given as 15 m/s^2 to the right.
Now, let's resolve the forces along the vertical and horizontal directions:
In the vertical direction, we have Ncos(θ) balancing the gravitational force mg:
Ncos(θ) = mg.
In the horizontal direction, we have Nsin(θ) providing the net force for the system:
Nsin(θ) = ma.
Since the mass is stationary relative to the wedge, the net force in the horizontal direction is 0. Therefore, we have:
Nsin(θ) = 0.
Since the mass does not move vertically, we have:
Ncos(θ) = mg.
Dividing the two equations, we get:
tan(θ) = (Nsin(θ))/(Ncos(θ))
tan(θ) = (0)/(mg)
tan(θ) = 0.
The only solution to this equation is θ = 0 degrees.
Therefore, the angle of inclination of the wedge is 0 degrees.
(b) To find the magnitude of the normal force exerted on the mass by the incline, we can use the vertical force equation we derived earlier: Ncos(θ) = mg.
Plugging in the values, we have:
Ncos(0) = (2.0 kg)(9.8 m/s^2).
Since cos(0) = 1, we can simplify the equation to:
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 magnitude of the normal force would increase. This can be seen from the equation Ncos(θ) = mg. As the acceleration increases, the normal force needs to be larger to balance the gravitational force.
Additionally, if the wedge were given a greater acceleration, the mass may start moving upwards or downwards depending on the angle of inclination. This is because the acceleration of the wedge would no longer cancel the gravitational force completely, resulting in a net force along the incline. The direction of this net force depends on the angle of inclination and the direction of the acceleration. Therefore, by increasing the acceleration, the mass would either move upwards or downwards along the incline.