Find the vertical component and friction

[DaveSauce]

Alright, so I've scanned my book and I am thuroughly confused.

A block of mass 6.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown in Figure P5.46. The coefficient of static friction between the block and the wall is 0.207. Determine the possible values for the magnitude of P that allow the block to remain stationary.

My friend has an identical problem (the numbers are different), and we can't even start working on the problem.

Of course, I'm not looking for an answer. I'm just trying to get started here...

what I'm thinking is that I need to find the vertical component of the P vector to be equal to first gravity, and second the vertical component involving friction...then use that to find the vector at 50 degrees, and thus find the horizontal component and then find the magnitude of the entire vector...

Some things that are confusing me. First, the normal vector. Is it pointing to the right, or is it vertical? Second, the friction vector...from what I gather, since the motion of the force P is upwards, the friction is supposed to oppose that and thus would point downwards with gravity.

Lastly, what equation do I even begin with? My professor in lecture today went over what friction is, how to find it using normal vector and the coefficient of friction...but he didnt even come close to mentioning how to apply it to problems. The previous problems I managed to find similar examples in the book for, but the book doesn't appear to mention anything like this problem. Normally I would ask the instructor, but the problems are web-based and due by 8am tomorrow...which is before anyone has recitation. I got all the other problems right, this is the last one I'm working on...

anyway, I attached a picture of the exact problem...I know the text is illegible, that's why I included the text above...

I'd like to thank in advance anyone who offers their help.

 

[paul11273]

I will try to get you started here.

First, to answer you question regarding the normal.
Normals are always perpendicular to the surface. If you are sitting in your chair reading this, the force of gravity (pulling down) is what keeps you in the chair. The normal force(pushing up from the chair) is what keeps you from falling to the floor. In you problem, the N force is acting from the right to the left. It is what keeps the book stopped at the wall, not continuing through it.

Second, the friction will be acting opposite of gravity, so up. Gravity is trying to pull the book toward the earth, and friction between the wall and block is opposing this.

The applied force P is pushing in an upward direction.

Draw your free body diagram with P up to the right at 50 degrees. Show your weight (mg) pulling down, and your friction going up.

From this you can figure your x and y components of each force. I will assume you use a standard cartesian axis with x on the horizontal and y on vertical. Since gravity is acting straight down, it has only a negative y component, the friction only positive y, remember friction is opposing gravity. I will leave the applied force components to you ( hint: x=rcos(theta) and y=rsin(theta) ).

Good luck.

 

[M98Ranger]

First, let's start by breaking down the problem into smaller parts. We have a block with a mass of 6.00 kg, pushed against a wall by a force P at a 50.0° angle with the horizontal. We also know that the coefficient of static friction between the block and the wall is 0.207.

To begin, let's draw a free body diagram of the block. This will help us visualize the forces acting on the block. From your description, it seems like you have a good understanding of the forces involved. The normal force will be perpendicular to the wall and the friction force will be parallel to the wall, in the opposite direction of the applied force P.

Next, let's look at the vertical forces. We know that the block is not moving, so the forces in the vertical direction must be balanced. This means that the vertical component of the applied force P must be equal to the weight of the block, which is given by mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). This can be written as Psin(50.0°) = mg.

Now, let's look at the horizontal forces. Again, since the block is not moving, the forces in the horizontal direction must also be balanced. This means that the horizontal component of the applied force P must be equal to the friction force. This can be written as Pcos(50.0°) = μN, where μ is the coefficient of friction and N is the normal force.

We now have two equations with two unknowns (P and N). We can solve for N by substituting the first equation into the second equation, giving us Pcos(50.0°) = μ(Psin(50.0°)/cos(50.0°)). Simplifying this, we get P = μtan(50.0°).

Now, we can plug this value of P back into the first equation to solve for N. We get Psin(50.0°) = mg, which can be rewritten as μtan(50.0°)sin(50.0°) = mg. Solving for N, we get N = mgsin(50.0°)/cos(50.0°).

To find the possible values for P, we need to consider the maximum value of P that will allow the block to