A uniform wooden plank 68cm long and weighing 38N rest horizontaly on two knife edges each 10cm from the ends of the plank. (1) What is the greatest land which applied at one end of the plank will not cause it to over turn. (2) what are the reactions of the support cohese a weight of 5N is supported at one end and another weight of 25N supported at the middle.
for balance:clockwise moment=anticlockwise moment:using one edge of knife as reference: L*10=38*24 L=91.2N (2)R1+R2=38+25+5=68N using one support:clockwise movement=48R1+5*10 anticlockwise moment=(25+38)*24 clockwise moment equals anticlockwise moment:48R1+50=1512 R1=33.6N R2=68-33.6 R2=34.4N
To solve this problem, we can consider the rotational equilibrium of the plank. The condition for equilibrium is that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Let's start by calculating the weight of the wooden plank. The weight of the plank is given as 38N.
Now, let's address the first question: "What is the greatest load which applied at one end of the plank will not cause it to overturn?"
To prevent the plank from overturning, the sum of the anticlockwise moments must be greater than or equal to the sum of the clockwise moments.
Let's assume the load applied at one end of the plank is W (the weight we need to find).
The clockwise moment is given by: 38N x (10cm) = 380Ncm
The anticlockwise moment is given by: W x (68cm - 10cm) = W x 58cm
To prevent overturning, the anticlockwise moment must be greater than or equal to the clockwise moment: W x 58cm ≥ 380Ncm
Now we can solve for W:
W ≥ 380Ncm / 58cm
W ≥ 6.55N
Therefore, the greatest load that can be applied at one end of the plank without causing it to overturn is 6.55N.
Now, let's move on to the second question: "What are the reactions of the supports when a weight of 5N is supported at one end and another weight of 25N is supported at the middle?"
In this case, we have two weights acting on the plank: 5N at one end and 25N at the middle.
Let's consider the reactions at the left support (supporting the 5N weight). We can use the principle of moments to find the reaction.
Clockwise moment = 38N x (10cm) = 380Ncm
Anticlockwise moment = 5N x (68cm - 10cm) = 5N x 58cm = 290Ncm
To find the reaction at the left support, which is the weight acting upwards, we set the sum of the moments to zero:
380Ncm = 290Ncm + reaction x (68cm - 10cm)
90Ncm = reaction x 58cm
reaction = 90Ncm / 58cm
reaction ≈ 1.55N
Therefore, the reaction at the left support for the 5N weight is approximately 1.55N.
To find the reaction at the right support (supporting the 25N weight), we can use the same principle of moments:
Clockwise moment = 38N x (68cm - 10cm) = 1756Ncm
Anticlockwise moment = 25N x (68cm / 2) = 850Ncm
To find the reaction at the right support, we set the sum of the moments to zero:
1756Ncm = 850Ncm + reaction x (68cm / 2)
906Ncm = reaction x 34cm
reaction = 906Ncm / 34cm
reaction ≈ 26.65N
Therefore, the reaction at the right support for the 25N weight is approximately 26.65N.
To summarize:
1) The greatest load that can be applied at one end of the plank without causing it to overturn is 6.55N.
2) The reaction at the left support for the 5N weight is approximately 1.55N.
3) The reaction at the right support for the 25N weight is approximately 26.65N.