There's a bar with length of 6.49m and mass of 5.45kg leaning against a frictionless wall. It makes an angle of 66.08 deg with the ground. A block with mass of 44.98 kg hangs from bar at distance d up from point of contact between bar and floor. The frictional force (us=.371) exerted by the floor keeps the bar from slipping. Find max d that the hanging mass can be attached to bar such that the bar does not slip.
second part: now the hanging mass is detached and removed from the problem. Find minimum angle between bar and ground for which the bar will not slip.
This is my first time on one of these sites and I'm really struggling with physics. will anyone help?
OK. In this, you have to use the conditions for equalibrium.
1) Sum of all vertical forces is zero. Remember at the wall, no vertical force.
2) Sum of all horizontal forces is zero, so you have friction at the bottom, and the wall force at the top.
3) sum of all moments about any point is zero. I would choose the bottom contact point, and sum there. Remember, moment is distance*force*sinTheta where theta is the angle between the distance and force.
These equations will solve for all the unknown forces (friction, wall force, etc).
Sure! I'd be happy to help you with your physics problem. Let's break it down step by step.
First, let's analyze the situation to understand what forces are acting on the bar. We have the weight of the bar acting downwards, the normal force exerted by the floor perpendicular to the surface of the bar, and the frictional force exerted by the floor parallel to the surface of the bar.
To find the maximum distance, d, that the hanging mass can be attached to the bar without causing it to slip, we need to consider the torque equilibrium condition. In other words, the sum of the torques acting on the bar must be zero.
The torque is calculated by multiplying the force acting on the bar by the perpendicular distance from the pivot. In this case, the pivot point is the point of contact between the bar and the floor.
Let's denote the distance from the pivot point to the point of attachment of the hanging mass as x. The torque due to the hanging mass will be the weight of the hanging mass multiplied by this distance: T_hanging_mass = m*g*x, where m is the mass of the hanging mass and g is the acceleration due to gravity.
We also need to consider the torque due to the weight of the bar itself. Since the bar is in equilibrium, the total torque due to the bar's weight must be zero. This torque is equal to the weight of the bar multiplied by half of its length since the weight force can be considered to act at the center of mass of the bar: T_bar = (1/2)*m_bar*g*L_bar, where m_bar is the mass of the bar and L_bar is its length.
Now, let's consider the torque due to the frictional force. The frictional force acts at the bottom of the bar, which is the point of contact with the floor. The torque due to friction can be calculated as: T_friction = f_friction*d_friction, where f_friction is the frictional force and d_friction is the distance from the pivot point to the point of application of the frictional force. In this case, d_friction is equal to half of the length of the bar since the frictional force acts at the bottom of the bar.
Finally, to ensure that the bar does not slip, the torque due to the hanging mass must be equal to the torque due to the frictional force, that is: m*g*x = f_friction*(L_bar/2).
Now, we can solve this equation for x to find the maximum distance that the hanging mass can be attached to the bar without causing it to slip.
To solve the equation, we need to substitute the values given in the problem:
- m = 44.98 kg (mass of the hanging mass)
- g = 9.8 m/s² (acceleration due to gravity)
- L_bar = 6.49 m (length of the bar)
- us = 0.371 (coefficient of static friction)
By substituting these values into the equation, we can solve for x.
Once the hanging mass is removed, we can consider the minimum angle between the bar and the ground for which the bar will not slip. In this case, the torque due to the hanging mass is zero, and thus, the torque due to the frictional force must also be zero. This means that we need to find the minimum angle for which the equation m*g*L_bar*sin(theta) = f_friction*(L_bar/2) is satisfied, where theta is the angle between the bar and the ground.
By rearranging the equation and substituting the give values, we can solve for the minimum angle theta.
I hope this explanation helps you understand how to approach and solve this physics problem. If you have any further questions, feel free to ask!