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Sum of Upward Forces Physics Problems

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Assignment - Questions and Answers

LO2

● Task1 :

A beam PQ is 5.0 m long and is supported at its ends in a horizontal

position as shown in the figure below. Its mass, a uniformly distributed

load, can be considered as equivalent to a single force of 400 N acting at

its centre, as shown. A point load of 12 kN acts on the beam in the position

shown. When the beam is in equilibrium, determine the reactions of the

supports.

- Solution :

Sum of upward forces = Sum of downward forces i.e.

+ R

= 12kN + 400N

+ R

= 12.4kN ...eq 1.1

Now, calculate moment about point Q

Distance of Load 12kN from Q = 2.5m + 1.3m = 3.8m = D

Distance of Centre of Gravity of beam from Q = 2.5m = D

Distance of P from Q = 5m = D

Thus, since beam is in equilibrium, moments are balanced,

i.e.

sum of moments in clockwise direction = sum of moments in anticlockwise

direction

x D

= 12 x 10

x D

+ 400 x D

x 5 = 12 x 10

x 3.8 + 400 x 2.5

Thus, we get R

= 9.32 kN

From eq 1.1, we calculate R

= 3.08 kN

= 9.32 kN

= 3.08 kN

● Task 2a:

A body weighs 2.8 N in air and 1.9 N when completely immersed in water of

density 1000 kg/m

. Calculate the volume of the body. Take the

gravitational acceleration as 9.81 m/s

- Solution :

weight of body in air = 2.8 N

weight of body in water = 1.9 N

density of water = 1000 kg/m

gravitational acceleration = g = 9.81 m/s

buoyant force on body = weight in air – weight in water = weight of volume

of water displaced

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But, volume of water displaced = volume of body submerged

Thus, weight in air – weight in water = density of water x volume of water

displaced x gravitational acceleration

i.e. 2.8 N – 1.9 N = 1000 x volume of body x 9.81 N

solving, we get:

volume of body = 9.1743x 10

-5

i.e. volume of body = 91.743 cm

Volume of body = 91.743 cm

● Task 2b:

A rectangular watertight box is 550 mm long, 410 mm wide and 205 mm deep. It

weighs 225 N. If it floats with its sides and ends vertically in water of

density 1030 kg/m

, what depth of the box will be submerged?

Take the gravitational acceleration as 9.81 m/s

- Solution :

Dimensions of box = 550 x 410 x 205 mm

Weight of box = 225 N

Density of water = 1030 kg/m

Gravitational acceleration = 9.81 m/s

Let submerged depth be ‘h’ mm

Since sides and vertical ends are submerged,

Volume of box submerged = volume of water displaced = h x 410 x 205 mm

Assume that the box is not completely submerged, i.e. h < 550 mm

buoyant force on body = weight in air – weight in water = weight of volume

of water displaced

since box is floating, weight of box in water = 0 N

thus, weight in air – weight in water = density of water x volume of water

displaced x gravitational acceleration

I.e.

225 N = 1030 x (h x 410 x 205) x 10 -9 x 9.81 N

solving, we get:

I.e.

h = 264.934 mm

Depth of the submerged part of box = 264.934 mm

● Task 3a:

The length of an iron steam pipe is 25.0 m at a temperature of 19 °C.

Determine the length of the pipe under working conditions when the

temperature is 290 °C. Assume the coefficient of linear expansion of iron is

12 ×10

-6

-1

- Solution :

Original length = L = 25.0 m

Temperature t

= 19°C

Temperature t

= 290°C

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Assignment - Questions and Answers LO2 ● Task1 : A beam PQ is 5.0 m long and is supported at its ends in a horizontal position as shown in the figure below. Its mass, a uniformly distributed load, can be considered as equivalent to a single force of 400 N acting at its centre, as shown. A point load of 12 kN acts on the beam in the position shown. When the beam is in equilibrium, determine the reactions of the supports. - Solution : Sum of upward forces = Sum of downward forces i.e. Rp + RQ = 12kN + 400N Rp + RQ = 12.4kN ...eq 1.1 Now, calculate moment about point Q Distance of Load 12kN from Q = 2.5m + 1.3m = 3.8m = DLQ Distance of Centre of Gravity of beam from Q = 2.5m = DCQ Distance of P from Q = 5m = DPQ Thus, since beam is in equilibrium, moments are balanced, i.e. sum of moments in clockwise direction = sum of moments in anticlockwise direction RP x DPQ = 12 x 103 x DLQ+ 400 x DCQ RP x 5 = 12 x 103 x 3.8 + 400 x 2.5 Thus, we get RP = 9.32 kN From eq 1.1, we calculate RQ = 3.08 kN RP = 9.32 kN RQ = 3.08 kN ● - Task 2a: A body weighs 2.8 N in air and 1.9 N when completely immersed in water of density 1000 kg/m3 . Calculate the volume of the body. Take the gravitational acceleration as 9.81 m/s2. Solution : weight of body in air = 2.8 N weight of body in water = 1.9 N density of water = 1000 kg/m3 gravitational acceleration = g = 9.81 m/s2 buoyant force on body = weight in air – weight in water = weight of volume of water displaced But, volume of water displ ...
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