Calculate Distance from Right End When Plank Tips

[Mcdeltateta]

1. Given a wooden plank mass 7 kg total length 4m supported by 2 sawhorses which are located l1=0.45m from left end and l2=1.5m from right end ...a cat with mass 3.9kg jumps on left end of plank and walks along to right side . How far is it from the right when plank begins to tip ?

Homework Equations

The Attempt at a Solution

 

[PhanthomJay]

1. Given a wooden plank mass 7 kg total length 4m supported by 2 sawhorses which are located l1=0.45m from left end and l2=1.5m from right end ...a cat with mass 3.9kg jumps on left end of plank and walks along to right side . How far is it from the right when plank begins to tip ?

Homework Equations

The Attempt at a Solution

I don't see your attempt.

 

[ogb p]

First, we need to calculate the center of mass of the plank and the cat. The center of mass is the point where the entire mass of an object can be considered to be concentrated. For a uniform object like the plank, the center of mass is simply the midpoint of the object.

The center of mass of the plank can be calculated as follows:

x_cm = (l1 + l2)/2 = (0.45m + 1.5m)/2 = 0.975m

This means that the center of mass of the plank is located 0.975m from the left end.

Next, we need to calculate the center of mass of the plank and the cat combined. This can be done by taking into account the masses of both objects and their respective distances from the center of mass.

m1x1 + m2x2 = (m1 + m2)x_cm

Where m1 and x1 are the mass and distance of the plank, and m2 and x2 are the mass and distance of the cat.

Substituting in the values given in the problem, we get:

(7kg)(0.975m) + (3.9kg)(0m) = (7kg + 3.9kg)x_cm

6.825m + 0 = 10.9kgx_cm

x_cm = 6.825m/10.9kg = 0.625m

Therefore, the center of mass of the plank and the cat combined is located 0.625m from the left end.

Now, we can use the concept of torque to calculate the distance from the right end when the plank begins to tip. Torque is the product of force and distance, and it causes an object to rotate. In this case, we can consider the weight of the plank and the cat as the force, and their respective distances from the right end as the distance.

Since the plank and the cat are in equilibrium, the sum of the torques acting on them must be equal to zero.

T_plank + T_cat = 0

Where T_plank is the torque due to the weight of the plank and T_cat is the torque due to the weight of the cat.

T_plank = (7kg)(9.8m/s^2)(1.5m) = 102.9Nm

T_cat = (3.9kg)(9.8m