Free body diagrams are your friends.

I get 18.85 for a and 26.32 for b. I'm going to double check Normal force will be shortened to just N Force of gravity = vertical component of normal force + vertical component of frictional force: 20=Ncos60 + .3Ncos30 Factor out N and you get N=20/(cos60+.3cos60), so 20/.7598=N The Horizontal force F can be solved by making the horizontal components equal, so Horizontal component of normal force = horizontal force of frictional force + horizontal force F Nsin60=.3Nsin30+F Substitute 26.32 for N and you get 22.80=3.95+F F=18.85

Is the answer meat bicycle?

AP physics 1 eh? If you can't understand this, good luck later this year. I took it my sophomore year and got a 4 on the AP test.

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I dropped outta school so if that dont matter...then yeah

We just went over that in my AP physics class. Fun stuff.

Oh come now guy? How hard is the object being pushed? How much mass does the object have? How much friction is the slope providing? How much gravity is the object under? It gave you all those numbers i assume. With those you can determine if the object can be pushed up the slope and exactly how fast. (Force of push must exceed the tendancy to fall back of the object + friction provided by the slope. If it does you can determine how fast up the slope the object can be pushed pending the ratio of force applied to object vs friction plus mass& gravity resistance.)

Lmao no

Is this for ap physics 1 or 2?

Pretty damn simple man. Equilibrium=box no move. Four things you need to figure out: Weight of box (W): m(mass of box) • g (acceleration due to gravity) Normal force (Fn): W • sin(60) Force of friction (f): Fn•us (static coefficient of friction) Minimum magnitude of applied force created equilibrium (F): W• cos(60) - f A) F = 5.1 B) Fn = 17 Took me longer to write this than it did to figure out those values. These problems are super easy if you split it up into component vectors. We know the mass of the box, so automatically we can figure out the force gravity is doing, as F = ma. From their, we know the normal force is perpendicular to the surface of the incline, so if we are to set the incline as 0° along the x-axis, the normal force is going in the direction of the y-axis, or 90°. Simple geometry tells us that the angle of the incline is the same as the angle the weight creates with from the y-axis on our arbitrary plane. So thus, we treat weight as the hypotenuse, and the normal force is weight • the sin of the incline's angle. Maximum force of static friction (be careful, you may not always be using the maximum, however since the box is in equalibrium, and we are looking for the minimum force to keep it there, we want the maximum static friction force.) is just the normal force • static coefficient. Now be conscious that the weight is not perpendicular to the incline, so that means there is an x-component of the weight that influences the box. Just like we found the normal force, but for the other side, weight • cos of the incline's angle. Now, since the box is not moving we need the forces on either x-side to cancel out. Since the x component of weight will be in the direction of the only force that would be doing work on the box if the applied force wasn't there, it will be in the direction of the motion. So we must set the forces going in the negative x direction (weight x-component) equal to the forces going in the positive x direction (friction and the applied force). Wx= f + F Since we need the applied force Wx - f = F.

I haven't even taken physics, but I'll take a shot. If it's at equilibrium, I assume that means that it isn't moving. This means that its acceleration is zero, so the net force is zero. This means that F is force pushing it up the hill, which is equal in magnitude to the force down the hill. The force down the hill is equal to the force of gravity minus the friction. You would then have to figure out the force of gravity, knowing that the slope is 60 degrees and a straight fall would be 90 degrees. The mass is 2 kg, and the acceleration is 9.8m/s,/s, so the force would be 19.6. Then I think you need to multiply by 60/90 or 2/3, which would get you about 13.1. Subtract the .3 from friction and F would be about 12.8 (I'm not sure what the unit is.) If this is wrong, it's because I haven't taken physics and I'm pretty much guessing. However, if this is right, then this problem isn't very hard and you probably need to pay more attention in class. EDIT: because acceleration is quadratic, maybe the correction for the angle should be 3600/8100, or 4/9. This would make the force of gravity 8.7, subtract the friction and F = 8.4. Or, if you square after the division, this would be 2/9, making the force of gravity 4.4, making F = 4.1. Again, I have no idea if this is correct or not. I might be missing some sort of constant.

The answer is 42.

Sorry to busy not doing my homework

Post a photo I'll have a look aha

no pics?