# Ice block on inclined plane- easy problem

• cep
In summary, a block of ice of mass 80kg is held stationary on a frictionless ramp at an angle of 36.9º above the horizontal. When held in place by a tangential force along the surface of the ramp, the magnitude of the force is 470 N. However, when held in place by a horizontal force directed towards the center of the ice block, the magnitude of the force is 590 N, contrary to the incorrect answer of 400 N provided in the book.
cep
Ice block on inclined plane-- easy problem!

I'm pretty sure I did this correctly, but my answers are different from those given in the back of the book (which our TA warned us were frequently incorrect), so I thought I'd ask you guys if I made any obvious mistakes.

## Homework Statement

A large block of ice of mass M=80kg is held stationary on a frictionless ramp. The ramp is at an angle of theta=36.9º above the horizontal.

a) If the ice block is held in place by a tangential force along the surface of the ramp (at angle theta to the horizontal), find the magnitude of the force.

b) If, instead, the ice block is held in place by a horizontal force, directed horizontally towards the center of the ice block, find the magnitude of this force.

Newton's laws

## The Attempt at a Solution

a) I rotated my axes by 36.9º so that the x-axis is parallel to the slope of the incline. Only the x-components matter in this problem, so I'm going to ignore the y-axis on here. For no acceleration to take place, F-mgx=0; F=mgx. In this case, mgx=mgsin(theta). F=80*9.8*sin(36.9)=470 N. The book's answer is 500 N, which is the same answer with 1 sigfig.

b) Keeping the same rotated axes, a horizontally applied force is not parallel to the axes. So, the relevant component of the force is Fx=Fcos(theta). Again, Fx-mgx=0 for no acceleration to take place. Fcos(theta)-mgsin(theta)=0

Fcos(theta)=mgsin(theta)
F=mgtan(theta)=80*9.8*tan(36.9)=590 N. The book's answer is 400 N, which doesn't make much logical sense to me, but maybe you all see where they're coming from. Thank you!

-CEP

cep said:
Fcos(theta)=mgsin(theta)
F=mgtan(theta)=80*9.8*tan(36.9)=590 N. The book's answer is 400 N, which doesn't make much logical sense to me, but maybe you all see where they're coming from.
The book's wrong, but you are correct. (I guess this is one of those times the TA warned you about.)

Thanks, Doc!

## 1. How does the angle of the inclined plane affect the movement of the ice block?

The angle of the inclined plane affects the force of gravity acting on the ice block. The steeper the angle, the greater the force of gravity and therefore the faster the ice block will slide down the plane.

## 2. What other factors besides angle can affect the movement of the ice block?

Friction between the ice block and the inclined plane can also affect the movement. If the surface of the plane is rough, it will cause more friction and slow down the ice block's movement.

## 3. What is the formula for calculating the acceleration of the ice block on the inclined plane?

The formula for calculating acceleration on an inclined plane is a = gsinθ, where a is the acceleration, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the inclined plane.

## 4. How does the mass of the ice block affect its movement on the inclined plane?

The mass of the ice block does not affect its movement on the inclined plane, as long as the force of gravity remains constant. This is because the mass of an object does not affect its acceleration, only the force acting on it does.

## 5. Is this problem considered an easy or difficult problem in physics?

This problem is considered an easy problem in physics because it only involves basic concepts such as gravity, acceleration, and friction. It does not require advanced mathematical calculations or complex theories to solve.

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