Physics' Questionn

Question

 

Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 53.0 N and F2 = 28.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

 

(image link http://www.webassign.net/CJ/04_11.gif )

Answer 1

Hello again. So we're looking for horizontal acceleration. That means we need to find all the forces in the horizontal plane. We have one which is F₂=28.5N but we can't say what the other force's horizontal component is yet.

To figure it out, we must first draw a right triangle for F₁ in such a manner that the force is the hypothenuse and 65º is to the right of the right angle. From this we can see that we need to use SohCahToa to find the horizontal component, more specifically, Cah for cosθ=adj/hyp. Adj is what we're looking for and we have the rest so our equation looks like this:

cosθ=adj/hyp

cos(65º)=adj/53N. Multiply each side by 53N to get:

53Ncos(65º)=adj

22.4N=adj

Now we know that our horizontal component of F₁ is 22.4N. Let's have this force facing the positive direction and F₂ face the negative direction. To know which way the block moves we must add the two forces:

F₁+F₂=22.4N-28.5N (since F₂ is facing the negative direction, it has a negative sign)

Our total force is -6.1N. So now we know that the magnitude is towards the left, the direction F₂ is being applied. To find the acceleration we now have to divide this force by the mass of the block (recall the explanation for this in your previous question):

-6.1N/5kg=-1.22m/s² and that's the answer. =]