# Forces On An Incline Homework Hotline

1) Since the system is in equilibrium, the vector sum of the forces acting on the block is 0. The forces acting on the block are gravity, force from the string (with the magnitude equal to the tension in the string, T), and the normal force. The normal force is perpendicular to the incline and pointing up and is balanced out by the component of the gravity perpendicular to the incline, pointing down. The force from the string on the block is balanced out by the component of the gravity parallel to the incline:

`T = mgsin(theta)`

So the tension in the string equals `mgcos(theta)` .

2) The length of the part of the string along the incline, l, can be found from the right triangle formed by the incline and the horizontal. Since `sin(theta)` = opposite/hypothenuse,

`sin(theta) = h/l`

From here `l = h/sin(theta)`

Then the length of the whole string is `L =h+h/sin(theta)` .

3) The mass per unit length is

`m/L = m/(h+h/sin(theta)) = (msin(theta))/(hsin(theta) + h)`

4) The speed of the waves in the string is given by

`v = sqrt(T/(m/L))`

Plugging in the mass per unit length and tension results in

`v = sqrt((mgsin(theta))/((msin(theta))/(hsin(theta) + h)))`

`v = sqrt(gh(sin(theta) + 1))`

5) Since the standing wave is set up only in the vertical part of the system, the lowest frequency standing wave will be

`f = v/(2h) = sqrt(gh(sin(theta) + 1))/(2h)`

`f = 1/2sqrt(g/h (sin(theta) + 1))`

6) If the standing wave has 3 nodes, its frequency is 4 times greater, so it will be

`2sqrt(g/h (sin(theta) + 1))`

Then, the period of this wave is `T = 1/f = 1/2sqrt(h/(g(sin(theta) +1)))`

7) The wavelength of this wave will be `lambda = vT` , or

`lambda = sqrt(gh(sin(theta)+1))*1/2sqrt(h/(g(sin(theta)+1)))`

`lambda = h/2`

The wavelength would be h/2, as expected for the standing wave with 3 nodes in the string of length h.

1. A 3.0 m long board has one end raised to a height of 60 cm to form an incline. A 4.0 kg mass is allowed to slide without friction down the entire length of the inclined plane.

a. What is the final speed of the mass when it reaches the bottom ?

b. If the mass is replaced with an 8.0 kg mass, what would be the new speed when it reaches the bottom?

2. A wooden block slides directly down an inclined plane, at a constant velocity of 6 m/s.

a. How large is the coefficient of kinetic friction if the plane makes an angle of 25 degrees with the horizontal?

b. If the angle of incline is changed to 10º, how long far will the block slide before coming to a stop?

3. A 10.0 kg box accelerates at 2.00 m/s^{2} as it slides down a ramp that makes an angle of 25.0 degrees with the horizontal. Find the coefficient of friction.

4. A 10.0 kg box is pulled up a 45 degree ramp at a constant velocity by a force of 90.0 N acting parallel to the ramp. Find the coefficient of friction.

5. a 10.0 kg box is pulled up a ramp that is at an angle of 20.0 degrees with the horizontal, with the rope that makes and angle of 30.0 degrees with the ramp. if pulling with a force of 70.0 N causes and acceleration of 2.00 m/s^{2}. Find the coefficient of friction.

6. A 10.0 kg box is pulled down a 45 degree ramp by a rope attached to the box that makes a angle of 30.0 degrees with the ramp. If a tension of 10.0 N causes the box to move at a constant velocity, what is the coefficient of friction for the box sliding on the ramp?

7. A car, with a mass of 1100kg, can accelerate on a level road from rest to 21 m/s in 14.0 s. What is the steepest slope this car can climb?

8. 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10 degrees with the horizontal. If there is no friction between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp?

9. A 10 kg block is placed on top of an inclined plane 10 m long. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2. The inclined plane is 30 degrees from the horizontal. a.) What is the acceleration of the block? b.) When will it hit the ground? c.) What is its impact velocity? d.) What is the angle of repose?

10. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and requires a force of 835 N parallel to the surface of the ramp to move it at a constant speed. What is the weight of the object?

11. The coefficient of static friction between a box and an inclined plane is 0.35. What is the minimum angle required for the box to begin sliding down the incline?

12. A package slides down a 135 m long ramp with no friction. If the package starts from rest at the top and is to have a speed no faster than 19m/s at the bottom, what should be the maximum angle of inclination?

13. A box of mass (m) is released from rest near the top of a frictionless sphere of radius, R. At what point does the box leave the surface of the sphere?

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