Question #54360

Here is a force diagram of the situation:

We can use this to come up with statements of the net force. Let's define up the ramp as the positive direction.

#sumF_"x"=f_s-F_"Gx"=ma_x#

#sumF_"y"=n-F_"Gy"=ma_y#

We know that there is no acceleration in the y-direction (the block does not move up or down relative to the surface of the ramp), so we have:

#n=F_"Gy"#

Using trigonometry, we see that the x-component of the gravitational force #F_"Gx"# is found using the sine function (see diagram).

#sin(theta)=(F_"Gx")/F_G#

#=>F_"Gx"=F_Gsin(theta)#

And similarly #F_"Gy"=F_Gcos(theta)#

Then we know that #F_G=mg#, so we may write:

#color(blue)(F_"Gx"=mgsin(theta))#

#color(blue)(F_"Gy"=mgcos(theta))#

Then:

#f_s-mgsin(theta)=ma_x#

Since we've established that there is no acceleration in the y-direction, all acceleration must be in the x-direction. We can solve for #a_x#:

#a_x=(f_s-mgsin(theta))/m#

We also know that #f_"smax"=mu_sn#, or #f_"smax"=mu_smgcos(theta)#, since we established above that #n=F_"Gy"#. Substituting this in, we get:

#a_x=(mu_smgcos(theta)-mgsin(theta))/m#

Then we can factor out an cancel #m#:

#a_x=(cancelm(mu_sgcos(theta)-gsin(theta)))/cancelm#

#=>a_x=mu_sgcos(theta)-gsin(theta)#

#=>color(blue)(a_x=g(mu_scos(theta)-sin(theta)))#

We are given:

• #mu_s=1/sqrt3#
• #theta=30^o#

So we can now solve for #a_x#:

#a_x=(9.81"m"//"s"^2)(1/sqrt3cos(30^o)-sin(30^o))#

Where #cos(30^o)=sqrt3/2# and #sin(30^o)=1/2#

#=>a_x=(9.81"m"//"s"^2)(1/sqrt3*sqrt3/2-1/2)#

#=>a_x=(9.81"m"//"s"^2)(1/2-1/2)#

#=>a_x=0" m"//"s"^2#