One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 60.2o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
F1L1-F2L2=0
L2= dsin(theta)
so,
d=(F1L1)/(F2sin(theta)
d= (2x1)/(6sin(60.2))
To find the position along the stick where the 6.00 N force is applied, we need to use the concept of torque. Torque is the rotational equivalent of force and is calculated as the product of the force and the distance from the pivot point (in this case, the pinned end).
Since we know that the net torque is zero, we can set up an equation using the torques caused by both forces to find the position of the 6.00 N force.
The torque (τ) caused by a force (F) is given by the equation:
τ = F * d
where d is the distance from the pivot point to the line of action of the force.
Let's label the distance from the pivot point to the 6.00 N force as x. The distance from the pivot point to the free end of the stick is 1.00 m (as the stick is a meter stick). We can write the equation for the torque caused by the 2.00 N force and the torque caused by the 6.00 N force as follows:
τ1 = (2.00 N) * (1.00 m)
τ2 = (6.00 N) * x
Since the net torque is zero, we have:
τ1 + τ2 = 0
(2.00 N) * (1.00 m) + (6.00 N) * x = 0
Simplifying, we have:
2.00 N + 6.00 N * x = 0
Rearranging the equation to isolate x:
6.00 N * x = -2.00 N
x = -2.00 N / 6.00 N
x = -0.333 m
The negative sign indicates that the 6.00 N force is applied on the opposite side of the pivot point from the 2.00 N force. Therefore, the 6.00 N force is applied at a distance of 0.333 m towards the pinned end of the stick.
So, the 6.00 N force is applied 0.333 m away from the pinned end of the stick.
To solve this problem, we can use the concept of torque and balance it out to find the position of the second force along the stick.
1. Firstly, let's label the following information:
- Distance from the pinned end to the point of application of the 2.00 N force: d1
- Distance from the pinned end to the point of application of the 6.00 N force: d2
- Angle between the 6.00 N force and the length of the stick: θ = 60.2°
2. Now, we know that torque is calculated by multiplying the magnitude of the force by the distance from the pivot point. To balance the torques, the clockwise torque must be equal to the counterclockwise torque.
3. Let's start by calculating the torque due to the 2.00 N force. Since this force is applied perpendicular to the length of the stick, the torque will be given by the product of the force magnitude and the distance from the point of application to the pivot:
Torque1 = (2.00 N) * d1
4. Next, let's calculate the torque due to the 6.00 N force. Since this force is at an angle, we need to calculate the perpendicular component of the force that acts along the length of the stick:
Perpendicular component of the 6.00 N force = (6.00 N) * cos(θ)
Now, the torque due to this force will be given by the product of the perpendicular component of the force and the distance from the point of application to the pivot:
Torque2 = (Perpendicular component of the 6.00 N force) * d2
So, the torque equation becomes:
Torque1 = Torque2
5. Substituting the values:
(2.00 N) * d1 = (6.00 N) * cos(θ) * d2
6. Finally, solving for d2, the distance from the pinned end to the point of application of the 6.00 N force:
d2 = (2.00 N * d1) / (6.00 N * cos(θ))
Now, you can substitute the known values of d1 and θ into this equation to find the value of d2, the distance along the stick where the 6.00 N force is applied.