A handbag containing some load weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is pulled \( 60^{\circ} \) to the vertical, find the force on each student's arm
81N
121N
162N
324N
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note : we where asked to look for the force in each student hand , not the total force in the both hand , if we where asked to look for the total force in both hand , the ans could have been162N , but since we where asked to look for the force in each hand we now solve like dis 
To find the force on each student's arm when they pull the handles of the bag at an angle of 60 degrees to the vertical, we can use trigonometry.
Given:
- Weight of the load = 162 N
- Angle between the handle and the vertical = 60 degrees
Let's denote:
- Force on each student's arm = F
- Vertical component of the force = F_vertical
- Weight of the load = W = 162 N
The force on each student's arm is the sum of the vertical components of the weight of the load. The vertical component can be calculated using trigonometry:
F_vertical = W * cos(angle)
1. Calculate the vertical component of the weight of the load:
F_vertical = 162 * cos(60)
2. Since each student is holding one handle, the force on each student's arm is equal to the vertical component of the weight of the load:
Force on each student's arm = F = F_vertical
3. Calculate the force on each student's arm:
F = 162 * cos(60)
By calculating the above expression, the force on each student's arm when they pull the handles at a 60-degree angle to the vertical is approximately 81 N.
My school sorry if my words are offensive but can't u solve principle of triangle. This isn't the first mistake or should I call 
it error.
This answer is wrong the the calculation is saying that the vertical component is fv=cos 60 and that is wrong vertical component is fv=sin60
