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Sum of Upward Forces Physics Problems

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Physics
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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
p
+ R
Q
= 12kN + 400N
R
p
+ R
Q
= 12.4kN ...eq 1.1
Now, calculate moment about point Q
Distance of Load 12kN from Q = 2.5m + 1.3m = 3.8m = D
LQ
Distance of Centre of Gravity of beam from Q = 2.5m = D
CQ
Distance of P from Q = 5m = D
PQ
Thus, since beam is in equilibrium, moments are balanced,
i.e.
sum of moments in clockwise direction = sum of moments in anticlockwise
direction
R
P
x D
PQ
= 12 x 10
3
x D
LQ
+ 400 x D
CQ
R
P
x 5 = 12 x 10
3
x 3.8 + 400 x 2.5
Thus, we get R
P
= 9.32 kN
From eq 1.1, we calculate R
Q
= 3.08 kN
R
P
= 9.32 kN
R
Q
= 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
3
. Calculate the volume of the body. Take the
gravitational acceleration as 9.81 m/s
2
.
- Solution :
weight of body in air = 2.8 N
weight of body in water = 1.9 N
density of water = 1000 kg/m
3
gravitational acceleration = g = 9.81 m/s
2
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
m
3
i.e. volume of body = 91.743 cm
3
Volume of body = 91.743 cm
3
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
3
, what depth of the box will be submerged?
Take the gravitational acceleration as 9.81 m/s
2
.
- Solution :
Dimensions of box = 550 x 410 x 205 mm
3
Weight of box = 225 N
Density of water = 1030 kg/m
3
Gravitational acceleration = 9.81 m/s
2
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
3
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
K
-1
.
- Solution :
Original length = L = 25.0 m
Temperature t
1
= 19°C
Temperature t
2
= 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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