1. ## physics 2

1.A man carries a 3 m long plank which has a mass of 3 kg over his shoulder. The tip of the plank extends 1 m in front of his shoulder. Given that his hand rests 0.3 m in front of his shoulder on the plank and that the weight of the plank is evenly distributed,
a) how much force does he need to exert in order for the plank to remain in the horizontal position over his shoulder?
b) what is the force exerted by his shoulder on the plank in this position?

2. Originally Posted by daphnewoon
1.A man carries a 3 m long plank which has a mass of 3 kg over his shoulder. The tip of the plank extends 1 m in front of his shoulder. Given that his hand rests 0.3 m in front of his shoulder on the plank and that the weight of the plank is evenly distributed,
a) how much force does he need to exert in order for the plank to remain in the horizontal position over his shoulder?
b) what is the force exerted by his shoulder on the plank in this position?
HI

Start by drawing the diagram , which i think you have done so and this is what you should have in your diagram .

There are three forces here , one acting downwards through the centre of mass (It's right in the middle since it says weight of the rod is uniformly distributed ) , the other one is the reaction of his shoulder on the rod which acts upwards , and the last one would be the force he exerts downwards to keep the rod in equilibrium .

Lets call the centre , G , the point where the rod touches his shoulder , P .

(a) By taking moments about P , 3(9.81)(0.5)=F(0.3)

Calculate F

(b) Taking moments about G , R(0.5)=F(0.3+0.5)

Here , use the F you calculated above

OR

force going upwards = force going downwards .

R=3(9.81)+F