A box of mass 21 kg sits on an inclined surface with an angle of 10 degrees. What is the component of the weight of the box along the surface?

Guest May 21, 2015

#5**+10 **

Melody, it is the "mg" vector that constitutes the hypotenuse of the right-angled triangle, not the "mgsinα" line (I didn't attempt to draw the two magnitudes to scale; I probably should have done!).

To convince yourself that the component along the slope is mgsinα think of the limit when α = 0. The component along the "slope" must then be zero.

.

Alan
May 25, 2015

#1**+5 **

21sin10. The force component along the slope is always mg sin theta where theta is the angle of the slope,m is the mass of the object and g is the force of gravity.

Mathcad
May 24, 2015

#4**+5 **

Thanks Alan but I am having problems with this :/

I can see that the two angles labelled theta in the first my pic are equal.

and

i can see using the second theta in my pic that

sin(theta)=weight down slope/mg so

**weight down slope = mg*sin(theta)**

BUT

If I use the same angle which you have labeled alpha then

sin(alpha)=mg/weight down the slope

**weight down the slope = mg/sin(alpha)**

Since I can see that these 2 angles are indeed equal I am totally confused by these 2 different outcomes.

What is wrong with my logic :/

Melody
May 25, 2015

#5**+10 **

Best Answer

Melody, it is the "mg" vector that constitutes the hypotenuse of the right-angled triangle, not the "mgsinα" line (I didn't attempt to draw the two magnitudes to scale; I probably should have done!).

To convince yourself that the component along the slope is mgsinα think of the limit when α = 0. The component along the "slope" must then be zero.

.

Alan
May 25, 2015