A bar with a mass of 20kg with a length of 30m is being lifted by a string and makes an angle of 55 degrees. The string is attached to the ceiling vertically.
What is the tension on the string?
What is the normal force given by the bar on the string
what is the force of static friction given by the ground on the bar?
Please proofread.
Where does the string attach to the rod?
What on earth do you mean by normal force on a string? Strings are usually in tension.
55 degrees with what?
*30cm bar
*55 degrees above the horizontal between the rod and the ground
*normal force given by the bar and the ground
*string is attached to one end of the string while other end is on the ground
sum of moments is zero. summing moments about the origin at the ground
20kg*9.8*15cm -Tension*length*sin(90-55)=0
that assumes the string is vertical at the point "length". if on the end, length-30cm.
Solve for tension.
<what is the normal force given by the bar on the string?> the negative of tension.
<static friction?>
The bar will tend to rotate about where the string is attached.
Summing momenmts about that point:
static -friction*length*sin55+verticalforceonground*length*cos55 + lenght/2*20*9.8=0 (moments in clockwise are negative, counterclockwise +)
Vertical force on ground=20kg*9.8-Tension
solve for friction
To find the tension on the string, we can break down the forces acting on the bar:
1. Weight force (mg): The weight force is vertically downwards and can be calculated by multiplying the mass of the bar (20kg) by the acceleration due to gravity (9.8 m/s^2). Hence, the weight force is 20 kg * 9.8 m/s^2 = 196 N.
2. Tension force (T): The tension force in the string keeps the bar from falling. It acts vertically upwards. To find its magnitude, we need to first resolve the weight force into its components. The vertical component, which is equal to the tension force, can be found by multiplying the weight force by the sine of the angle (55 degrees) between the weight force and the vertical axis. Therefore, T = 196 N * sin(55 degrees) = 161.12 N.
So, the tension on the string is approximately 161.12 N.
To find the normal force exerted by the bar on the string, we need to consider the equilibrium of forces perpendicular to the bar's length:
3. Normal force (N): The normal force acts perpendicularly to the surface upon which the bar rests. In this case, it is the tension force that provides the vertical component of the normal force. Therefore, the normal force is equal to the tension, which is approximately 161.12 N.
So, the normal force given by the bar on the string is approximately 161.12 N.
To find the force of static friction given by the ground on the bar, we need to consider the equilibrium of forces parallel to the surface upon which the bar rests:
4. Force of static friction (fs): The force of static friction opposes the tendency of the bar to slide. In this case, since the bar is not sliding, the force of static friction is equal in magnitude but opposite in direction to the force applied on the bar (which is the horizontal component of the tension force). The horizontal component of the tension force can be found by multiplying the tension force by the cosine of the angle (55 degrees) between the weight force and the vertical axis. Therefore, fs = -T * cos(55 degrees) = -161.12 N * cos(55 degrees) = -105.16 N (negative sign indicates direction).
So, the force of static friction given by the ground on the bar is approximately -105.16 N.