# Solve Friction Problems: 3.00 kg Crate on 35° Incline

• 404
In summary, the coefficient of static friction between a 3.00 kg crate and a 35 degree incline is 0.300. To prevent the crate from sliding down the incline, a minimum force of 7.23N perpendicular to the incline must be applied. In the second conversation, a truck carrying a load (20,000 kg) with a coefficient of static friction of 0.500 on the truck bed and moving at 12 m/s is discussed. To prevent the load from sliding forward, the net force must be 0, meaning that the acceleration must also be 0. This can be achieved by increasing the normal force, which also increases the force of friction.
404
The coefficient of static friction between the 3.00 kg crate and the 35 degree incline is 0.300. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I managed to do most of the question(whether it's wrong or right), Figured out friction was 7.23N, and the box slides down with 9.64 N if the force is not applied... But what I'm confused on is, how do you use a Y component to cancel out the x component? (I mean if you draw a coordinate plane perpendicular to the incline, then the 9.64 is going straight down the x-axis and the perpendicular froce being asked is going straight down the y axis)

Truck (10,000 kg) carries a load (20,000 kg), moving at 12 m/s. If load is not tied down and coefficient of static friction is 0.500 with truck bed, calculate the minimum stopping distance for which the load will not slide forward relative to the truck. And is there any unless info?

I figure if the load can't slide forward, the net force should be 0, therefore accerlation must be 0 too, but that's where I'm stuck...

Force of friction is mu times the normal force. So if you increase the normal force, what happens to the force of friction?

ahh right friction increases too, so that's how you cancel out the x component of gravity... Thanks, I'll go try it now.

## 1. What is friction and how does it affect objects on an incline?

Friction is a force that opposes the motion of an object. On an incline, it acts in the direction opposite to the object's motion, making it more difficult for the object to move.

## 2. How do I calculate the frictional force on a 3.00 kg crate on a 35° incline?

The frictional force can be calculated using the formula F = μN, where μ is the coefficient of friction and N is the normal force acting on the object. To find N, we need to break down the weight of the crate into its components parallel and perpendicular to the incline.

## 3. What is the normal force acting on the crate?

The normal force is the force exerted by the incline on the crate, perpendicular to its surface. In this case, it is equal to the component of the crate's weight perpendicular to the incline, which can be calculated using trigonometry.

## 4. How does the coefficient of friction affect the motion of the crate on the incline?

The coefficient of friction is a measure of the roughness between two surfaces in contact. A higher coefficient means that there is more resistance to motion, making it more difficult for the crate to move. A lower coefficient means less resistance and easier motion.

## 5. How can I reduce the friction and make it easier for the crate to move on the incline?

To reduce friction, you can use a lubricant or make the surfaces smoother. You can also decrease the weight of the crate or decrease the angle of the incline. Additionally, changing the material of the surfaces in contact can also affect the coefficient of friction.

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