Practice
practice problem 1



solution
For all solutions, let T_{1} be the cable on the left and T_{2} be the cable on the right. The sign always has weight (W), which points down. The sign isn't going anywhere (it's not accelerating), therefore the three forces are in equilibrium. Describe this state using the language of physics — equations; in particular, component analysis equations. As always, make a nice drawing to show what's going on. Use a ruler and a protractor if you wish.
 The two upward components should equal one another. Together they should equal the weight, which means each one is carrying half the load.
T_{1} = T_{2} = ½W
T_{1} = T_{2} = ½(100 N)
T_{1} = T_{2} = 50 N  Weight points down (270°) and T_{1} points to the left (180°). These are both good vectors — good in the sense that they are easy to deal with. T_{1} is the troublemaker. Break it up into components and state the conditions for equilibrium in the vertical and horizontal directions. I like to put negative vectors on the left side of the equals sign and positive vectors on the right side. I also suggest working through the vertical equation first.
horizontal ∑F_{−x} = ∑F_{+x} T_{1} = T_{2} cos θ_{2} T_{1} = 141 N cos 45° T_{1} = 100 N vertical ∑F_{−y} = ∑F_{+y} W = T_{2} sin θ_{2} T_{2} = W = 100 N sin θ_{2} sin 45° T_{2} = 141 N  Weight is the only force with a convenient direction. Resolve the tensions into their components. State the equilibrium condition along both axes. I suggest working with the horizontal equation first.
horizontal ∑F_{−x} = ∑F_{+x} T_{1} cos θ_{1} = T_{2} cos θ_{2} vertical ∑F_{−y} = ∑F_{+y} W = T_{1} sin θ_{1} + T_{2} sin θ_{2} That's the end of the physics. The rest of the work is math. Solve the horizontal equation for T_{1}. Substitute the result into the vertical equation.
T_{1} = T_{2} cos θ_{2} cos θ_{1} W = T_{2} cos θ_{2} sin θ_{1} + T_{2} sin θ_{2} cos θ_{1} Solve that for T_{2}, substitute values, and compute T_{2}.
T_{2} = W (sin θ_{1} cos θ_{2})/(cos θ_{1} + sin θ_{2}) T_{2} = 100 N (sin 60° cos 30°)/(cos 60° + sin 30°) T_{2} = 50.0 N Substitute back into the horizontal equation and compute T_{1}.
T_{1} = T_{2} cos θ_{2} cos θ_{1} T_{1} = 50.0 N cos 30° cos θ_{1} T_{1} = 86.6 N
My, that last one wasn't very much fun. Let's see if there isn't a simpler solution. We used component analysis since it's the default approach. Whenever you're given a pile of vectors and you need to combine them, components is the way to go — especially if you have no expectation of any special relationships among the vectors. We use this brainless, brute force approach to problems all the time. Understand the rules, describe them using commands a computer understands, put numbers in, get answers out.
Sometimes, however, there are clever solutions available. They don't work all the time, but when they do we should use them. In this practice problem, the vectors are rigged so that the alternate solution is easier than the default solution. The graphical method for addition of vectors requires placing them head to tail. The sum would be the resultant vector connecting the tail of the first vector to the head of the last. When forces are in equilibrium, their sum is zero and their will be no resultant. This means, it should be possible to arrange the three vectors in this practice problem into a closed figure — a triangle. Let's try it.
 This is what we call a degenerate triangle. Sure it has three sides, but it covers no area. The two short sides lie on top of the long side. Symmetry tells us the two short sides should have equal length. Thus each tension equals half the weight. We already said this, so there is no advantage to this method over the previous one.
T_{1} = T_{2} = ½W
T_{1} = T_{2} = ½(100 N)
T_{1} = T_{2} = 50 N  The horizontal tension and the vertical weight are the legs of a 45–45–90 triangle whose hypotenuse is the diagonal tension. These forces should form the ratio 1:1:√2.
T_{1} = W = T_{2} 1 1 √2 T_{1} = 100 N = T_{2} 1 1 √2 T_{1} = 1(100 N) T_{1} = 100 N T_{2} = √2(100 N) T_{2} = 141 N  The two tensions are the legs of a 30–60–90 triangle and weight is the hypotenuse. This means the sides should form the ratio 1:√3:2. Just be sure to get the tensions to correspond to the correct parts of this ratio.
T_{2} = T_{1} = W 1 √3 2 T_{2} = T_{1} = 100 N 1 √3 2 T_{1} = √3 100 N 2 T_{1} = 86.6 N T_{2} = 1 100 N 2 T_{2} = 50 N
And there are other ways to solve this problem.
practice problem 2
 What is the net force?
 What fourth force will put the point in equilibrium?
solution
Solution…
 Compute the x and y components of each vector. Arrange the results in a table like this one…
# magnitude direction x component y component 1st 3 N 0° +3 N +0 N 2nd 4 N 90° +0 N +4 N 3rd 5 N 217° −4 N −3 N resultant equilibrant Add the components…
# magnitude direction x component y component 1st 3 N 0° +3 N +0 N 2nd 4 N 90° +0 N +4 N 3rd 5 N 217° −4 N −3 N resultant total +1 N +1 N equilibrant Use pythagorean theorem to get the magnitude of the resultant force…
∑F = √(F_{x}^{2} + F_{y}^{2}) ∑F = √((1 N)^{2} + (1 N)^{2}) ∑F = √(2 N^{2}) ∑F = 1.41 N Use tangent to get the direction…
tan θ = F_{y} = 1 N = 1 F_{x} 1 N θ = 45 ° Add these numbers to the table…
# magnitude direction x component y component 1st 3 N 0° +3 N +0 N 2nd 4 N 90° +0 N +4 N 3rd 5 N 217° −4 N −3 N resultant 1.41 N 45° +1 N +1 N equilibrant  The fourth force that would put this arrangement in equilibrium (the equilibrant) is equal and opposite the resultant. The components work this way too. To get the opposite direction angle, add on 180°.
# magnitude direction x component y component 1st 3 N 0° +3 N +0 N 2nd 4 N 90° +0 N +4 N 3rd 5 N 217° −4 N −3 N resultant 1.41 N 45° +1 N +1 N equilibrant 1.41 N 135° −1 N −1 N
practice problem 3
 Draw a free body diagram of the crate.
 If the angle of the ramp is set to 10°, determine…
 the component of the crate's weight that is perpendicular to the ramp
 the component of the crate's weight that is parallel to the ramp
 the normal force between the crate and the ramp
 the static friction force between the crate and the ramp
 At what angle will the crate just begin to slip?
solution
Solution…
 Draw it
[insert free body diagram]
 This is an example of a classic physics problem that students have been solving since the Seventeenth Century.
 The component of the crate's weight perpendicular to the ramp is found using the cosine function. An object's weight is entirely pushing into a surface when the surface is level (a 0° angle of inclination). None of that weight is pushing into the surface when the surface is vertical, like a wall (a 90° angle of inclination). Cosine is a maximum when the angle is zero and zero when the angle is 90°. This is how the perpendicular component works.
W_{⊥} = W cos θ = mg cos θ W_{⊥} = (100 kg)(9.8 m/s^{2})(cos 10°) W_{⊥} = 96.5 N  The component of the crate's weight parallel to the ramp is found using the sine function. An object's weight has no sideways component on a level floor (a floor with no inclination). An object's weight is entirely parallel to a wall (a floor with a 90° inclination, in a sense). Sine is zero when the angle is zero and a maximum when the angle is 90°. This is how the parallel component works.
W_{∥} = W sin θ = mg sin θ W_{∥} = (100 kg)(9.8 m/s^{2})(sin 10°) W_{∥} = 17.0 N  Normal forces are normal — that is, perpendicular to a tangent drawn to a curve or surface. I'll say it again, this crate isn't currently going anywhere, so all the forces perpendicular to the incline must cancel. For a static crate on an incline, the force normal to the incline equals the perpendicular component of its weight.
N = W_{⊥} N = 96.5 N  Friction is a sideways or lateral force — a force that is parallel to the surface of a solid. This crate isn't going anywhere, so all the forces parallel to the incline should cancel. For a static crate on an incline, the static friction force equals the parallel component of the crate's weight.
ƒ = W_{∥} ƒ = 96.5 N
 The component of the crate's weight perpendicular to the ramp is found using the cosine function. An object's weight is entirely pushing into a surface when the surface is level (a 0° angle of inclination). None of that weight is pushing into the surface when the surface is vertical, like a wall (a 90° angle of inclination). Cosine is a maximum when the angle is zero and zero when the angle is 90°. This is how the perpendicular component works.
 The component of the crate's weight parallel to the incline pulls the crate down the incline while the frictional force tries to keep it in place. Since nothing is going anywhere, these two forces must balance each other.
∑ F = m a W_{∥} − ƒ = 0 ƒ = W_{∥} As the angle of inclination increases, so to does the static friction, but it can't keep doing this forever. At some angle, the parallel component of the weight will equal the maximum static friction. Friction won't be strong enough and the crate will slip.
ƒ_{max} = W_{∥} μ_{s}mg cos θ = mg sin θ Cancel the weight.
μ_{s} cos θ = sin θ
Do some trig.
tan θ = μ_{s}
Enter numbers.
tan θ = 0.28_{}
Compute. The angle at which the crate just begins to slip is…
θ = 16°
This number is known as the critical angle (because it marks a critical value separating two types of behavior — sticking vs. sliding), angle of friction (because you gotta call it something), angle of repose (because granular materials will settle, or repose, in conical piles with this angle), or critical angle of repose (because adding grains to a pile with this angle will make it slump).
 This part of the problem moves us from the Seventeenth to the Eighteenth Century, because it include Newton's second law of motion in one direction. When it come to components, the choice of trig functions are still true. Only the angles have been changed to protect the innocent.
 The perpendicular component follows the sames rules it did in part a. Use cosine here.
W_{⊥} = W cos θ = mg cos θ W_{⊥} = (100 kg)(9.8 m/s^{2})(cos 130°) W_{⊥} = 84.9 N  The parallel component of the weight still uses the sine function.
W_{∥} = W sin θ = mg sin θ W_{∥} = (100 kg)(9.8 m/s^{2})(sin 30°) W_{∥} = 49.0 N  There is no way for the crate to move perpendicular to the ramp in this scenario. The normal force must therefore equal the perpendicular component of the crate's weight.
N = W_{⊥} N = 84.9 N  The angle of inclination in part d is greater than the critical angle calculated in part c. Friction is no longer strong enough to keep the crate in place. The kinetic friction in this part of the problem is now really a function of the material surfaces (the coefficient of friction) and the contact forces (the normal force).
ƒ = µ_{k}N ƒ = (0.17)(84.9 N) ƒ = 14.4 N  The forces perpendicular to the surface cancel out. The forces parallel to the surface do not. One is greater than the other. The parallel component of the weight is greater than the kinetic friction force. The difference of these two is the net force, and it drags the crate down the ramp.
∑F = W_{∥} − ƒ ∑F = 49.0 N − 14.4 N ∑F = 34.6 N down the ramp  Welcome to the Eighteenth Century. Net force and mass determine acceleration. This is Newton's second law of motion — discovered in 1666, making it the most modern part of this problem.
a = ∑F m a = 34.6 N 100 kg a = 3.46 m/s^{2}
 The perpendicular component follows the sames rules it did in part a. Use cosine here.
practice problem 4
solution
Answer it.