Basic tension
![Picture](/uploads/6/1/5/8/61582483/published/swing.png?1492642733)
A \(20.0 N\) child sitting on a playground swing is being pushed by her father. When the rope makes an angle of \( 27^{\circ} \) to the vertical, what is the force exerted by her father? What is the tension in the rope, \( T \)?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( F_{g} = 20.0 N \)
- \( \theta = 27^{\circ} \)
- \( T = \text{?} \)
- \( F = \text{?} \)
USEFUL FORMULA(S)
- \( \text{sin} \theta = \frac{ \text{opp} }{ \text{hyp} } \)
- \( \text{cos} \theta = \frac{ \text{adj} }{ \text{hyp} } \)
- \( \text{tan} \theta = \frac{ \text{opp} }{ \text{adj} } \)
- \( a^{2} + b^{2} = c^{2} \)
PROCESS
- Rearrange the force vectors to form a triangle, with the known angle included.
- Using the standard trigonometric ratios, solve for \( T \).
- \( \text{cos} \theta = \frac{ \text{adj} }{ \text{hyp} } \)
- \( \text{cos} (27^{\circ}) = \frac{ F_{g} }{ T } \rightarrow T = \frac{ 20 N }{ \text{cos} (27^{\circ}) } \)
- \( \Rightarrow T = 22.44652475 N \Rightarrow T \approx 22.4 N \)
- Use the Pythagorean Theorem to solve for \( F \).
- \( a^{2} + b^{2} = c^{2} \)
- \( F_{g}^{2} + F^{2} = T^{2} \rightarrow F = \sqrt{ T^{2} - F_{g}^{2} } \)
- \( F = \sqrt{ 503.8 - 400 } \)
- \( \Rightarrow F = 10.19050899 N \Rightarrow F \approx 10.2 N \)
- The child's father is exerting a force of approximately \( 10.2 N \), and there are roughly \( 22.4 N \) of tension in the rope.
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with vector triangles. We grabbed the formulas for the Pythagorean Theorem as well as the standard trigonometric ratios.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the cosine ratio to solve for \( T \), as well as the Pythagorean equation needed to be modified slightly.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
FRICTION
![Picture](/uploads/6/1/5/8/61582483/published/table.png?1492642782)
A \( 15 kg \) object rests on a table. A cord is attached to this object and also to a wall. Another object is hung from this cord as shown. If the coefficient of friction between the \( 15 kg \) object and the table is \( 0.27 \), what is the maximum mass that can be hung, without movement?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( m_{1} = 15 kg \)
- \( \mu = 0.27 \)
- \( \theta = 30^{\circ} \)
- \( m_{2} = \text{?} \)
USEFUL FORMULA(S)
- \( F_{f} = \mu \cdot F_{n} \)
- \( F = ma \)
- \( \text{sin} \theta = \frac{ \text{opp} }{ \text{hyp} } \)
- \( \text{cos} \theta = \frac{ \text{adj} }{ \text{hyp} } \)
- \( \text{tan} \theta = \frac{ \text{opp} }{ \text{adj} } \)
PROCESS
- Rearrange the force vectors to form a triangle, with the known angle included.
- Using the friction formula, solve for \( F_{f} \)
- \( F_{n} = ma \)
- \( F_{n} = (15)(9.8) \)
- \( \Rightarrow F_{n} = 147 N \)
- \( F_{f} = \mu \cdot F_{n} \)
- \( F_{f} = (0.27)(147) \)
- \( \Rightarrow F_{F} = 39.69 N \)
- Using the standard trigonometric ratios, solve for \( F_{g} \)
- \( \text{tan} (30^{\circ}) = \frac{ F_{g} }{ F_{f} } \rightarrow \frac{ F_{g} }{ 39.69 } \)
- \( \Rightarrow F_{g} = 22.920800569 N \Rightarrow F_{g} \approx 22.9 N \)
- Then, use Newton's Second Law to solve for \( m \)
- \( F = ma \rightarrow m = \frac{F}{a} \)
- \( m = \frac{ 22.9 }{ 9.8 } \)
- \( \Rightarrow m = 2.33826859 kg \Rightarrow m \approx 2.34 kg \)
- The maximum mass that can be hung, without movement, is roughly \( 2.34 kg \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with tension and vector triangles. We grabbed the formulas for tension, Newton's Second Law, as well as all standard trigonometric ratios.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the cosine ratio to solve for \( F_{g} \), as well as Newton's Second Law to solve for \( m \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
ANgles
![Picture](/uploads/6/1/5/8/61582483/published/rope.png?1492645059)
A \( 675 N \) object is pulled horizontally by a force of \( 410 N \) as shown. What is the angle, \( \theta \), between the rope and the vertical?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( F_{g} = 675 N \)
- \( F = 410 N \)
- \( \theta = \text{?} \)
USEFUL FORMULA(S)
- \( \text{sin} \theta = \frac{ \text{opp} }{ \text{hyp} } \)
- \( \text{cos} \theta = \frac{ \text{adj} }{ \text{hyp} } \)
- \( \text{tan} \theta = \frac{ \text{opp} }{ \text{adj} } \)
PROCESS
- Rearrange the force vectors to form a triangle, with the known angle included.
- Using the standard trigonometric ratios, solve for \( \theta \).
- \( \text{tan} \theta = \frac{ F }{ F_{g} } \rightarrow \frac{ 410 N }{ 675 N } \)
- \( \text{tan}^{-1} ( \frac{ 410 }{ 675 } ) = \theta \)
- \( \Rightarrow \theta = 31.27480548^{\circ} \Rightarrow \theta \approx 31.3^{\circ} \)
- The angle between the rope and the vertical is roughly \( 31.1^{\circ} \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with angles and vector triangles. We grabbed the formulas for all the standard trigonometric ratios.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the tangent ratio to solve for \( \theta \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
uniform beam 1
![Picture](/uploads/6/1/5/8/61582483/published/bar.png?1492724042)
A student stands on a uniform \( 25 kg \) beam. The scale on the right end reads \( 350 N \). What is the mass of the student?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( m_{beam} = 25 kg \)
- \( d_{student} = 1.2 m \)
- \( d_{scale} = 3.0 m \)
- \( d_{beam} = 1.5 m \)
- \( F_{scale} = 350 N \)
- \( F_{beam} = \text{?} \)
- \( F_{student} = \text{?} \)
USEFUL FORMULA(S)
- \( \tau = Fr \)
- \( F = ma \)
PROCESS
- Solve for \( F_{student} \)
- \( \sum \tau = 0 \)
- \( \tau_{student} + \tau_{beam} = \tau_{scale} \)
- \( (F_{student})(1.2 m) + (245 N)(1.5 m) = (350 N)(3 m) \)
- \( (F_{student})(1.2 m) = 682.5 N \cdot m \)
- \( F_{student} = \frac{ 682.5 N \cdot m }{ 1.2 m } \)
- \( \Rightarrow F_{student} = 568.75 N \)
- Solve for \( m \).
- \( F_{student} = m (9.8) \)
- \( m = \frac{ 568.75 }{ 9.8 } \)
- \( \Rightarrow m = 58.03571429 kg \Rightarrow m \approx 58 kg \)
- The student has a mass of approximately \( 58 kg \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with torque. We grabbed the formulas for torque and Newton's Second Law.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the torque equations to solve for \( F_{student} \), as well as modify Newton's Second Law to solve for \( m \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
uniform beam 2
![Picture](/uploads/6/1/5/8/61582483/published/beam.png?1492724248)
A \( 35 kg \) uniform plank is balanced at one end by a \( 55 kg \) student as shown. What is the overall length of the plank?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
GENERAL STEPS
- \( m_{student} = 55 kg \)
- \( m_{beam} = 35 kg \)
- \( d_{student} = 1.3 m \)
- \( d_{beam} = \frac{ L }{ 2 } \)
- \( L = \text{?} \)
USEFUL FORMULA(S)
- \( \tau = Fr \)
- \( F = ma \)
PROCESS
- Calculate the \( F_{g} \) of the student and the beam.
- \( F_{student} = ( m_{student} )( 9.8 \frac{ m }{ s^2 } ) \)
- \( F_{student} = (55)(9.8) \)
- \( \Rightarrow F_{student} = 539 N \)
- \( F_{beam} = ( m_{beam} )( 9.8 \frac{ m }{ s^2 } ) \)
- \( F_{beam} = (35)(9.8) \)
- \( \Rightarrow F_{beam} = 343 N \)
- Solve for \( d_{beam} \).
- \( \sum \tau = 0 \)
- \( \tau_{student} = \tau_{beam} \)
- \( ( F_{student} )( d_{student} ) = ( F_{beam} )( d_{beam} ) \)
- \( (539)(1.3) = (343)( d_{beam} ) \)
- \( \frac{700.7}{343} = d_{beam} \)
- \( \Rightarrow d_{beam} = 2.042857143 m \Rightarrow d_{beam} \approx 2.0 m \)
- Solve for \( L \)
- \( d_{beam} + d_{student} = \frac{ L }{ 2 } \)
- \( 2.042857143 + 1.3 = \frac{ L }{ 2 } \)
- \( 2( 3.342857143 ) = L \)
- \( \Rightarrow L = 6.685714286 m \Rightarrow L \approx 6.7 m \)
- The overall length of the plank is roughly \( 6.7 m \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with torque. We grabbed the formulas for torque and Newton's Second Law.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the torque equations to solve for \( d_{beam} \), as well as the Pythagorean equation.
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
double tension
![Picture](/uploads/6/1/5/8/61582483/published/light.png?1492723936)
A \( 35 kg \) traffic light is suspended from two cables as shown in the diagram. What is the tension in these cables?
INFORMATION GIVEN
USEFUL FORMULA(S)
PROCESS
- \( m = 35 kg \)
- \( \theta _{1} = 140^{\circ} \)
- \( \theta _{2} = 100^{\circ} \)
- \( \theta _{3} = 120^{\circ} \)
- \( F_{g} = \text{?} \)
- \( T_{1} = \text{?} \)
- This is Cable 1.
- \( T_{2} = \text{?} \)
- This is Cable 2.
USEFUL FORMULA(S)
- \( \frac{ a }{ \text{sin} (A) } = \frac{ b }{ \text{sin} (B) } = \frac{ c }{ \text{sin} (C) } \)
- \( F = ma \)
- \( F_{g} = m(9.8) \)
PROCESS
- Calculate the force of gravity on the traffic light.
- \( F_{g} = (35)(9.8) \)
- \( \Rightarrow F_{g} = 343 N \)
- Break down the larger angles into smaller angles referring to the vertical axis.
- Rearrange the force vectors to form a triangle, with the known angles included.
- Use the Sine Law to solve for the missing sides:
- \( \frac{ T_1 }{ \text{sin} (60) } = \frac{ 343 N }{ \text{sin} (80) } = \frac{ T_2 }{ \text{sin} (40) } \)
- \( T_1 = \text{sin} (60) ( \frac{ 343 }{ \text{sin} (80) } ) \)
- \( \Rightarrow T_1 = 301.6291379 N \Rightarrow T_1 \approx 302 N \)
- \( T_2 = \text{sin} (40) ( \frac{ 343 }{ \text{sin} (80) } ) \)
- \( \Rightarrow T_2 = 223.8773501 N \Rightarrow T_2 \approx 224 N \)
- \( \frac{ T_1 }{ \text{sin} (60) } = \frac{ 343 N }{ \text{sin} (80) } = \frac{ T_2 }{ \text{sin} (40) } \)
- The tension in Cable 1 is roughly \( 302 N \), and the tension in Cable 2 is approximately \( 224 N \).
GENERAL STEPS
- Identify the information given by the problem, and the value(s) it's asking for.
- Identify what type of problem it is, and gather the relevant formula(s).
- In this case we're dealing with angles and vector triangles. We grabbed the formulas for the Sine Law and Newton's Second Law.
- Break down the useful formula(s) into the components of information available to you.
- Our formulas were already how we wanted them.
- Redraw the force vectors in the form of a triangle, maintaining all information (direction, magnitude, etc.) exactly as stated.
- Rearrange the equation(s), if necessary, until the unknown value is isolated.
- We had to rearrange the Sine Law to solve for both \( T_1 \) and \( T_2 \).
- Insert the known values into this new equation, and solve.
- Repeat until all unknown values are found, if needed!
THE CHALLENGE
This question had many different aspects to it, including vector triangles, angles, and tension. It was challenging to break down each vector into its proper sections and create an accurate vector triangle.
THE SOLUTION
The first thing that I tried to do was create a vector triangle. In order to do so, I had to break down each of the larger angles and find values for the unknown, smaller ones. Luckily, this seemed to be the correct course of action. After I had created my vector triangle, I saw that it didn't contain a right angle. Thus, I needed to bring in the formula for the Sine Law and solve for the unknown values. I already knew the "length" of one vector, being the gravitational force of the traffic light, and used that to find the "lengths" of the other cables via the angles I found earlier.