Question

A 2.00 meter long horizontal plank whose mass is 3.00 kg is placed so that the very end (the right end) rests on a retaining wall. The left end is put on the tailgate of the truck.
However the tailgate is tilted a bit, so it only contacts the board at a point 25.0 cm from the end. A 2.50 kg sledge hammer is placed 0.500 meters from the retaining wall.

a)Find the force acting on the board by the tailgate.

b)Find the force acting on the board by the retaining wall.

Answer 1

Answer:

$R_g=\frac{61}{7} =8.7143\ N$

$R_w=45.1857\ N$

Explanation:

Given that:

• mass of plank, $m=3\ kg$

• length of plank, $l=2\ m$

From the image we can visualize the given situation.

Consider the given plank to be mass-less and having a uniformly distributed mass of 1.5 kg per meter.

Now in the balanced condition:

• $\sum F=0$

$R_g+R_w=(3+2.5)\times 9.8$

$R_g+R_w=53.9\ N$ .......................(1)

• $\sum M_w=0$

$1.75\times R_g=3\times 1+(2.5\times 9.8)\times 0.5$

$R_g=\frac{61}{7} =8.7143\ N$ ...........................(2)

is the force acted by the tailgate on the plank.

Substitute the value from (2) into (1):

$R_w=45.1857\ N$ is the force acted by the wall upon the plank.