Recent content by th77

I attached my message and the problem in a Word document because I don't know how to enter equations in this text box. Thanks for your help

thanks for your help

You're right because I got the answer using it, but I don't know why. The problem says that only 2 'horizontal' forces act on the object. Why is the vertical involved? I know F2 is making an angle, but then what do they mean by horizontal forces? Shouldn't it just say 'forces'? I have a...

I think it tells me that it should be F1 - F2 instead. If yes then a = (9 - 3.76)/3 = 1.75 It still doesn't come to 2.9

Only two horizontal forces act on a 3.0 kg body. One force is 9.0N, acting due east, and the othr is 8N, acting 62 degrees north of west. What is the magnitude of the body's acceleration? Fnet= ma ...so I took that to mean F1x + F2x = ma F1x = 9 cos 0 = 9 f2x = 8 cos 62 = 3.76 a = (9 +...

Thanks very much!

The weight along the x-direction is mgsin 8 . Does it matter what the y component of weight is? Using Newtons 2nd law would it be T - mg sin 8 = ma ?

I'm not sure. Thats part of my question.

Holding on to a towrope moving parallel to a frictionless ski slope, a 50 kg skier is puilled up the slope, which is at an angle of 8 degrees with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at...

Thanks very much for you help!

No, down to the left. Would it be correct to say that acceleration is an abstract concept? Meaning that it has nothing to do with the direction of an object's motion.

So then they could've chosen positive x direction to be up the plane? Is there any advantage to choosing positive x to be down the plane? I'm guessing the initial velocity is negative because it's going up the incline plane, which here is the negative x direction. My concept of acceleration...

A block is projected up a frictionless plane with initial speed v0 = 3.50 m/s. The angle of incline is theta = 32 degrees. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it bets back to the bottom? I read through the solution...

the vector is in the 3rd quadrant so isn't it 240?

Thanks! That leands me to another question... In this problem, the angle is 30 degrees with the negative axis so shouldn't cos 240 be equal to sin 30? They come to -0.5 and 0.5 respectively.