Test: Vector Analysis and Forces Acting on an Object - 2 - MCAT MCQ

• If the clothes are hung in the middle of the line, this means we can treat the horizontal portion of a triangle as 3ft, and with the line sagging down 3ft, we can treat this as a 45-45-90 triangle, since the vertical and horizontal sides are equal.
• If the weight of the clothes pulls the line into the equivalent of a 45-45-90 triangle triangle, we can treat the force vectors accordingly.
• Because there are two anchor points for the clothesline, we can assume the horizontal components cancel each other out. The vertical component, 30N (F=ma, using 10m/s² for the gravitational constant) should then equal Tcos45 + Tcos45, or 2Tcos45 since we have a 45-45-90 triangle. (sin45 could also be used, remember that cos45 = sin45)
• Solve for T:
  2∗T∗cos 45 = 30
  T∗cos 45 = 15
  T∗(1/√(2) T = 15
  T = 15√(2)

• The tension in the tow rope is dependent only upon the weight of the lighter car. Therefore the force due to gravity would be 5,000N.
• On an inclined surface, the tension in the rope would be equal to the sinθ times the force vector hypotenuse, so in this case, we have a θ = 30° and the vector of force would be our force due to gravity of 5,000N.
• Our equation then would be 5,000 sin30 = T, where T is the tension in the rope. Sin30 is ½, therefore T is 2,500.
• If T is 2,500 then the rope can withstand 500N more force. Since tension in this example will be equal to ½ mg due to sin30 = ½ , and our T must be 500, we can use this formula solve for m.
  T = 1/2mg
  500N = ½m x 10m/s² 
  m = 100kg

If the angle is taken between the ceiling and the wire, then the vertical component would be given by the sin θ of the tension, but in this case, we are given the vertical component of force already.

The 4 wires horizontal components will negate one another, so the only vertical force involved is the force of gravity from the circus performer.

Since 4 separate wires are involved, your equation for the vertical component of a wire should be ¼ 60kg x 9.8 m/s² or 147 N.

• If there is a 60° angle to the ground, this gives us a 30-60-90 triangle.
• In this case, the horizontal component of force should be equal to the cos θ multiplied by the force of tension.
• Since we know θ = 60, and T = 30N, we can solve: 30 cos 60 = 15 N.

• Because the two wires are at different angles, they will bear different tensions, however, the horizontal components will remain the same. In this case, T1 cos 30 = T2 cos 60. This gives us √(3)/2 T1 = ½ T2. We can then multiply both sides by 2 to get:
• √(3) T1 = T2
• We know that the combined vertical component of tension must match the force of gravity from the sandbag, therefore, T1 sin 30 + T2 sin 60 = 300 N or ½ T1 + √(3)/2 T2 = 300 Multiply by 2 on both sides to get:
• T1 + √(3) T2 = 600.
• We can substitute √(3)T1 for T2 in our second equation due to what we calculated in the first, this gives us an equation of: T1 + √(3) (√(3)T1) = 600
• We can simplify this to T1+3T1= 600. Solving for T1 gives us a value of T1=150N

• A force of 5 N would be the net force in one direction, not the tension in the rope.
• While it may be intuitive to think that pulling on either end of the rope may result in the two opposing forces adding together, this is not the case. The tension in the rope is not 145N.
• Tension is determined by the weaker of two forces. A helpful way to think about it is considering that if only one side of the rope is pulled, the tension in the rope is zero. It’s only by having a weaker opposing force that tension is created.

Since we are given the mass, tension needed, and angles of suspension, and our variable is the number of wires. The vertical component would be Tsin45 x = 400, where x is the number of cables.

Since our maximum tension is 10 N, we can set T to 10 and solve: 10 √(2)/2 x = 400.

Rearranging the equation will give us x = 40√(2). Though you cannot use a calculator, you can automatically eliminate 20 and 40 as answers, because √(2) is greater than 1.

We also know that 40x2=80, and √(2) is less than 2, so we can eliminate 80 as a choice, since 60 is a smaller number while still being able to support the sign.

Since 60 = 1.5 x 40, and √(2) is less than 1.5, we can conclude that 60 cables would be sufficient to support the sign though technically only ~57 cables would be needed.

• In this case, only the first string is responsible for the vertical component of tension, while the second string, being parallel to the ground, only opposes the first string horizontally.
• Because the first string fully carries the vertical component, we can solve for the tension in the first string with the equation 50 N = T1sin30. This gives us T1=100 N.
• The tension in the second string only has a horizontal component. And since the two horizontal components must be equal, we can set T1cos30 = T2. Solve for T2:
  100 cos30 = T2
  100 √(3)/2 = T2
  50 √(3) = T2

• In this scenario the swing is acting as a simple pendulum, and the tension is acting against gravity.
• Since the swing is in periodic motion, the tension, on the rope would be greatest at the bottom of the period; when the swing is at its highest speed.
• To calculate tension, we would use the equation: T = mg cosθ
• Solving for this problem: T = mg cosθ
  T = 30∗10∗cos45
  T = 300∗(1/√2)
  T = 300/√2
  T = 150∗√2
  ​Remember to rationalize the denominator by multiplying both the numerator and denominator by √2.

• The tension in the rope can be found by the equation F_g sin30 = T
• Since the block is 30kg, this gives us a gravitational force of 300 N.
• Since the block is on a 30° slope, we multiply Fg by sin30 (½) to get a T of 150 N.