Cliff and Will are carrying a uniform 2.0m board of mass 75kg. Will is supporting the board at the end while cliff is 0.6m from the other end as shown in the following figure. Cliff has attached his lunch to his end of the board, and the tension in the string supporting the lunch is 190N. Find the normal forces exerted by Cliff and Will.
Can someone help with the formulas for these? Thank you.
190 down at left
C up at x = .6
75*9.81 down at x = 1
W up at x = 2
forces up = forces down
C + W = 190 + 75*9.81 = 926 Newtons
moments about x = 0 sum to 0
-.6 C + 75*9.81*1 - 2 W = 0
or 2W +.6C = 736 Newton meters
Thanks for your time! I don't understand this answer. Maybe someone else can help me understand the formulas to find the force for Cliff and the force for Will.
Thank you
C + W = 926 so
2 W = 1852 - 2 C
then
(1852 - 2 C) + .6 C = 736
1852 - 736 = 1.4 C
etc
Thanks again, Damon. Still not getting this at all. I appreciate your time, but the steps still don't make sense to me.
add up the forces. That is one equation
add up the torques (moments) about x = 0. That is the second equation
To find the normal forces exerted by Cliff and Will, we need to consider the forces acting on the board.
1. Weight of the board:
The weight of the board is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2). The formula for weight is:
Weight = Mass * Gravity
2. Weight of the lunch:
Similarly, the weight of the lunch is equal to its mass multiplied by the acceleration due to gravity.
3. Tension force:
The tension force in the string supporting the lunch is given as 190N. This force acts upwards.
4. Normal force at Cliff's end:
The normal force at Cliff's end is the force exerted by the board on Cliff. It acts perpendicular to the board, pushing upwards. To find this, we need to sum up the forces acting in the vertical direction. The formula for this can be written as:
Normal force at Cliff's end = Weight of board + Weight of lunch - Tension force
5. Normal force at Will's end:
The normal force at Will's end is the force exerted by the board on Will. It also acts perpendicular to the board, pushing upwards. To find this, we need to sum up the forces acting in the vertical direction. The formula for this can be written as:
Normal force at Will's end = Weight of board + Weight of lunch - Tension force
Now, let's find the numerical values:
1. Weight of the board:
Weight = Mass * Gravity
Weight = 75 kg * 9.8 m/s^2
2. Weight of the lunch:
Weight = Mass * Gravity
It is not given in the question. You need to provide the mass of the lunch.
3. Tension force:
Given in the question as 190N.
4. Normal force at Cliff's end:
Normal force at Cliff's end = Weight of board + Weight of lunch - Tension force
5. Normal force at Will's end:
Normal force at Will's end = Weight of board + Weight of lunch - Tension force
Once you have the values for the weight of the board, the weight of lunch, and the tension force, you can substitute them into the formulas to calculate the normal forces exerted by Cliff and Will.