# Thread: This triangle is on rollers and equilibrium. What is the force x? Thank you

2. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

Originally Posted by cxz7410123
Is this interesting question given to you to solve for yourself?
OR do you expect us to solve it for you?

If you post some work, then maybe we will know how to help you to a solution.

3. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

Actually my drawing was wrong. I have uploaded the correct one. Which basically is a hinge.

My solution is that since the object is in equilibrium, the horizontal 100N must be equal to x*sin20 -----> x=100/sin20 -----> x=292
The problem is that from my everyday experience, force x should be less than 100. For example when you close a door, the hinge slides in with less force than the spring which pushes it.
So either my calculation is wrong or my experience is wrong.
Any help?

4. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

Originally Posted by cxz7410123
Actually my drawing was wrong. I have uploaded the correct one. Which basically is a hinge.

My solution is that since the object is in equilibrium, the horizontal 100N must be equal to x*sin20 -----> x=100/sin20 -----> x=292
The problem is that from my everyday experience, force x should be less than 100. For example when you close a door, the hinge slides in with less force than the spring which pushes it.
So either my calculation is wrong or my experience is wrong.
Any help?
I think the value you found for x is correct, but you are thinking about it incorrectly. x is not the force of gravity. If gravity alone were acting on the system, it would begin accelerating to the right. The 100N force is the force to prevent it from accelerating to the right. I would expect that force to be more similar to the force that the hinge slides in with and the vertical force as the spring force (if you are still using the door comparison). So, if the system weighs 30kg, it would require a 100N force from the right to keep it from starting to roll. Again, this is my interpretation of the problem, so please forgive me if I am wrong.

5. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

I apologize for my English, it is not a hinge it is latch.

Ignoring gravity and friction, do you think that force x on it is almost 3 times (x=292) the force of a spring (100)?

6. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

Originally Posted by cxz7410123
I apologize for my English, it is not a hinge it is latch.

Ignoring gravity and friction, do you think that force x on it is almost 3 times (x=292) the force of a spring (100)?
Yes

7. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

I think I have made a mistake in my calculation. There must be an Fx and an Fy reaction to the 100N
Could you please check my new calculations?
(I accidentally changed θ from 20 to 25 but this is not the case)

8. ## Re: This triangle is on rollers and equilibrium. What is the force x? Thank you

Originally Posted by cxz7410123
I think I have made a mistake in my calculation. There must be an Fx and an Fy reaction to the 100N
Could you please check my new calculations?
(I accidentally changed θ from 20 to 25 but this is not the case)

No. That is not correct. The 100N force has only a horizontal component. It is pushing against a horizontal surface, so the reaction is 100N in the horizontal direction (same as the force applied). It is 0N in the vertical direction (same as the force applied).

When the force labeled x is pushing down on the slanted surface, the force acts on the object both horizontally and vertically. Imagine the vertical force as a finger. The flatter it is (closer to horizontal), the less horizontal force there will be. Since this is a relatively vertical surface, the larger the horizontal component will be. So, you are correct that your original thoughts were incorrect. If there is a 20 degree slant from the vertical, then there is a 70 degree slant from the horizontal. We were using the wrong trigonometry. The horizontal component of the reaction force is $x\sin 70^\circ$ is the force pushing to the right, 100 is the force pushing to the left, $x\cos 70^\circ$ is the force pushing down. Since the rollers are against the ground, the ground pushes back with equal force to whatever the downward force is, so that is guaranteed to be in equilibrium. The horizontal forces are in equilibrium when $x\sin 70^\circ = 100$ or when $x = \dfrac{100}{\sin 70^\circ} \approx 106.42\text{N}$